Math 112: Calculus I

Fall Term, 2004

InstructorDavid Murphy


Extra Problems

Each week I'll try to add a few interesting problems on this page for you to consider and then I'll post their solutions at the end of the week. This will give you, I hope, an additional resource as some of you requested, for challenging problems to try in preparation for the upcoming test and to have their solutions as well (eventually).


November 17, 2004

Optimization Worksheet from Nov. 17: problems and solutions

Problem 1:  A poster is to have an area of 180 in2 with 1 in margins at the bottom and on the two sides and a 2 in margin at the top. What dimensions give the largest printed area?

Solution:  Let x be the width of the poster and y be its height. Then we must have that xy = 180. What we want to maximize is the printed area, which will be A = ( x - 2 )( y - 3 ), where we subtract 2 from x since we have 1 in margins on each side and 3 from y since we have a 2 in margin at the top but a 1 in margin at the bottom. From the constraint that xy = 180, we have y = 180 / x, so we want to maximize the function

A(x) = ( x - 2 )( [ 180 / x ] - 3 ) = 180 - 3x - [ 360 / x ] + 6 = 186 - 3x - [ 360 x-1 ].
To find the dimensions that maximize the printed area, we find the critical points of A(x) (where x is potentially any number strictly larger than 0, since it can't equal zero for xy = 180 and it can't be negative as it is a dimension). Thus we consider
A ' (x) = -3 - [ -360 x-2 ] = -3 + 360 x-2 = [ -3 x2 + 360 ] / x2
The critical numbers of A(x) are those values of x where either A ' (x) = 0 or where A ' (x) does not exist. The only place where A ' (x) dne is when x = 0, but 0 is not in the domain of A, so our only critical number arises from the solution of A ' (x) = 0, which happens when -3x2 + 360 = 0, i.e., when 3x2 = 360 or x2 = 120. Thus x = +1201/2 is the only critical number of A(x) (because x > 0 is the domain of A). To confirm that A(x) is maximized when x is the square root of 120, consider the Second Derivative Test, for A " (x) = -720 x-3 implies that A " (+1201/2) < 0 in which case A(x) has a local maximum here. In fact, on the entire domain { x > 0 } of A(x), A " (x) = -720 x-3 < 0 so the curve is concave down on the entire domain and x = 1201/2 is the only place where A(x) could have a maximum, so this local maximum is in fact also an absolute maximum of the function. Therefore, the printed area of the poster is largest when the poster is x = 1201/2 = 10.95 inches wide and y = 180 / 1201/2 = 16.43 inches high.

Problem 2:  Your iron works has contracted to design and build a 500 ft3, square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding 1/2-in-think stainless steel plates together along their edges. As the production engineer, your job is to find the dimensions for the base and height that will make the tank weigh as little as possible. What dimensions do you tell the shop to use?

Solution:  The first thing to realize is that, as the thickness of the tank will be uniformly 1/2-in-thick and stainless steel's weight is proportial to its volume, the tank's weight will be as little as possible when we minimize the surface area of the tank. If x is the length of a side of the bottom of the tank (which is a square, so all four sides of the bottom have length x) and y is the height of the tank, then its volume is V = x2 y, which must equal 500 according to the contract. Thus y = 500 / x2 = 500 x-2. Again, we want to minimize the surface area of the open-top rectangular tank, which is

S = 4 xy + x2 = 4 x [ 500 x-2 ] + x2 = 2000 x-1 + x2,
since the tank has four sides each of area base times height equals xy and one bottom of area x2. Thus, to minimize S(x), we find its critical numbers and test them. So consider
S ' (x) = -2000 x-2 + 2x = 2 [ -1000 + x3 ] / x2
This is zero when x = 10 and does not exist when x = 0. However, x = 0 is not in the domain, for x2 y = 500 implies x = 0 is impossible (and x < 0 is not in the domain since x is a length), so the only critical number of S(x) is x = 10. To test whether this is a maximum or a minimum, use the Second Derivative Test, for S " (x) = 4000 x-3 + 2 is always > 0 for x > 0, which is the domain of S(x). In particular, S " (10) = 6 > 0 implies that S(x) has a local maximum whem x = 10, which is then an absolute maximum for S(x) since S is always concave down on its domain { x > 0 }. Therefore, you should tell your shop to make the tank so that its base is x = 10 ft by x = 10 ft, while its height should be y = 500 / 102 = 5 ft.

Problem 3:  The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 m apart, where should an object be placed on the line between them so as to receive the least illumination?

Solution:  The illumination of the object will be the sum of the illuminations of the two light sources, of strengths S1 and S2 = 3 S1 respectively. Let x be the distance between the object and the first light, so 10 - x is its distance from the second light source. Then

I(x) = I1(x) + I2(x) = k [ S1 / x2 ] + k [ ( 3 S1 ) / ( 10 - x )2 ] = k [ S1 x-2 ] + 3k S1 ( 10 - x )-2
is the total illumination of the object. For simplicity, let's denote the strength of the first light source by S and omit the subscript 1. The domain is { 0 < x < 10 } as we are placing the object between the two light sources. So let's find the critical numbers of I(x) in this domain:
I ' (x) = -2k S x-3 - 6k S [ ( 10 - x )-3 ( -1 ) ] = -2k S x-3 + 6k S ( 10 - x )-3 = 2k S [ -( 10 - x )3 + 3x3 ] / [ x3 ( 10 - x )3 ].
The critical numbers are where I ' (x) = 0 or I ' (x) dne. The latter only happens when x = 0 or x = 10, neither of which are in the domain { 0 < x < 10 } so the only critical numbers occur when I ' (x) = 0, which only happens when the numerator is 0. Since k and S are constants, the numerator is zero only when 3x3 = ( 10 - x )3, and taking cube-roots of both sides we find that 31/3 x = 10 - x, so ( 31/3 + 1 ) x = 10, so x = 10 / ( 31/3 + 1 ) = 4.09 meters. We still have to confirm that this is where the object receives the least illumination, so consider
I '' (x) = 6k S x-4 - 18k S ( 10 - x )-4
which has I '' (4.09) = 6k S [ (4.09)-4 - 3 (5.91)-4 ] = 6k S [ 0.0011145 ] > 0. Therefore, applying the Second Derivative Test, we conclude that I(x) has a local minimum when the object is placed 4.09 m from the weaker light source.

Problem 4:  Let v1 be the velocity of light in air and v2 the velocity of light in water. According to Fermat's Principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken, where C is a point on the water surface. Show that

v1 / v2 = sin q1 / sin q2
where q1 is the angle of incidence (the angle between AC and the normal to the water's surface at C) and q2 is the angle of refraction (the angle between CB and the normal to the water surface at C).

Solution:  Let h be the height of A above the water's surface, d be the depth of B beneath the water's surface, and D be the horizontal distance between A and B. All of these are constants. Let x be the horizontal distance between A and C, the point on the water's surface where the light should enter and bend to go down to B. According to Fermat's Principle, the time taken should be minimized, so let's figure out the time taken in terms of x and the constants h, d, and D. As x is the horizontal distance from A to C and D is the horizontal distance from A to B, D - x is the horizontal distance from C to B. By the Pythagorean Theorem, the distance from A to C is [ h2 + x2 ]1/2 while the distance from C to B is [ d2 + ( D - x )2 ]1/2. As light travels at velocity v1 in the air (i.e., on its path from A to C) and v2 in the water (i.e., from C to B), the time it takes light to travel from A to C is

tAC(x) = |AC| / v1 = [ h2 + x2 ]1/2 / v1
using the familiar equation from physics that distance = velocity times time, solved for time. Similarly, the time taken for the light to travel from C to B is
tCB(x) = |CB| / v2 = [ d2 + ( D - x )2 ]1/2 / v2
as above. Thus, the time taken for light to travel from A to B along the path from A to C to B is
t(x) = tAC(x) + tCB(x) = [ h2 + x2 ]1/2 / v1 + [ d2 + ( D - x )2 ]1/2 / v2
which is what we want to minimize. Thus, let's find its critical numbers on the domain { 0 < x < D } = [0,D]. So we consider
t ' (x) = ( 1/2 ) [ h2 + x2 ]-1/2 [ 2x ] / v1 + ( 1/2 ) [ d2 + ( D - x )2 ]-1/2 [ 2 ( D - x )1 [ -1 ] ] / v2 = [ x / ( v1 [ h2 + x2 ]1/2 ) ] - [ ( D - x ) / ( v2 [ d2 + ( D - x )2 ]1/2 ) ] = [ x ( v2 [ d2 + ( D - x )2 ]1/2 ) - ( D - x )( v1 [ h2 + x2 ]1/2 ) ] / [ v1 v2 [ h2 + x2 ]1/2 [ d2 + ( D - x )2 ]1/2 ].
The denominator is never zero, for the velocities v1, v2 are non-zero and the square roots refer to the distances |AC| > 0 and |CB| > 0 respectively (i.e., |AC| > h and |CB| > d regardless of the value of x). Therefore, the only critical number for t(x) occurs when t ' (x) = 0, which happens only when the numerator is zero. Thus,
x ( v2 [ d2 + ( D - x )2 ]1/2 ) = ( D - x )( v1 [ h2 + x2 ]1/2 )
so that
v1 / v2 = [ x / [ h2 + x2 ]1/2 ] / [ ( D - x ) / [ d2 + ( D - x )2 ]1/2 ] = sin q1 / sin q2
because x / [ h2 + x2 ]1/2 is the ratio of the side opposite the angle q1 (on the worksheet) and the hypotenuse, and likewise for sin q2. Notice that this is actually true at every critical number of t(x), so all that is left to show in order to prove Snell's Law is to ensure that t(x) is not minimized at either of the endpoints, x = 0 or x = D. When x = 0, t ' (0) = [ -D ( v1 h ] / [ v1 v2 h [ d2 + D2 ]1/2 ] < 0, so t(x) is decreasing at the left endpoint, so it doesn't have its absolute minimum when x = 0. Similarly, when x = D, t ' (D) > 0, so that t(x) was increasing up to its right endpoint, so it couldn't have had its absolute minimum when x = D eithere. Therefore, t(x) has its absolute minimum at one of its critical points, and we have shown above that at such points Snell's Law, v1 / v2 = sin q1 / sin q2, is true.

Problem 5:  You have been asked to design 1000-cm3 cans shaped like right circular cylinders.

  1. What dimensions will use the least material if we ignore the waste in manufacturing? That is, if we only worry about minimizing the surface area of the can, which is
    S = 2 p r2 + 2 p r h.
  2. What dimensions will use the least material if we take waste into account? (Assume there is no waste in cutting the material for the sides, but the tops and bottoms of radius r will be cut from squares that measure 2r cm on a side. The total amount of metal used by each can will therefore be
    A = 8 r2 + 2 p r h
    rather than S = 2 p r2 + 2 p r h as in (a).)

Solution:  In both parts, we have the same constraint that the volume of the can must be 1000 cm3, and the volume of a cylinder is p r2 h. Hence p r2 h = 1000, so h = 1000 / ( p r2 ) and we'll write S and A in terms of r only using this substitution and the domains of each will be { r > 0 }.

  1. We are asked to minimize S = 2 p r2 + 2 p r h = 2 p r2 + 2 p r ( 1000 / p r2 ) = 2 p r2 + 2000 r-1. Therefore, as always, let's find the critical numbers:
    S ' (r) = 4 p r - 2000 r-2 = 4 [ p r3 - 500 ] / r2,
    which is zero when r = ( 500 / p )1/3 = 5.42 and S ' (r) is always defined on the domain { r > 0 }. So r = 5.42 is the only critical number and S '' (r) = 4 p + 4000 r-3 is always greater than 4 p > 0 when r > 0 implies that S(r) has not only a local min when r = 5.42 cm, but in fact that this is when its absolute minimum occurs. Thus the can's dimensions should be r = 5.42 cm and h = 1000 / [ p ( 5.42 )2 ] = 10.84 cm.
  2. Now we want to minimize the function A(r) = 8 r2 + 2 p r [ 1000 / ( p r2 ) ] = 8 r2 + 2000 r-1. As above, consider
    A ' (r) = 16 r - 2000 r-2 = 16 [ r3 - 125 ] / r2,
    which is defined everywhere on its domain, { r > 0 }, but is zero when r3 = 125, i.e., when r = 5 cm. Then A '' (r) = 16 + 4000 r-3, which is always strictly greater than 16 > 0 when r > 0, so A '' (5) > 0 and the amount of material is therefore minimized when r = 5 cm and h = 1000 / [ p ( 5 )2 ] = 12.73 cm.
Notice that when we take waste into account our ideal radius decreases, which makes sense since the waste we're taking into account is from cutting the circles to make the ends of the cylinders, and these pieces only depend on r and not on h, so reducing r while increasing h allows us to decrease the size of the squares we need to use to cut ends from for our can.

Problem 6:  The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a rectangular box with a square end the largest possible volume?

Solution:  There are actually two cases to this problem, since "length" refers to the length of the longest side, which may or may not be one of the sides of the square base. We'll consider the cases separately below and then compare answers at the end. First, let's call the side of the square end x and the third dimension y, so the volume of the box is V = x2 y, and this is what we want to maximize. Also we observe that the Postal Service's constraint that length plus girth be < 108 may be replaced, for the sake of maximizing volume, by the constraint that length plus girth equal 108, for if the sum is some number smaller than 108, we could increase the value of y leaving x alone so that length plus girth is still < 108, but not V = x2 y will have increased (since we increased y). So when the volume is maximum, we must have that length plus girth is equal to 108.

So we have two situations in which volume was maximized. In the first, when (x = 18,y = 36), we have V = 11664. In the second, when (x = 24,y = 18), we had V = 10368. As 116644 > 10368, the volume of a box we may mail using the U.S. Postal Service is largest when it has dimensions 18" x 18" x 36".

Problem 7:  A 4 m length of wire is available for making a circle and a square. How should the wire be distributed between the two shapes to maximize the sum of the enclosed areas?

Solution:  Let x be the length of the wire used to construct the circle, which leaves 4 - x meters of wire with which to make the square. Now the x meters forms the circumference of the circle constructed, which means that x = C = 2 p r, where r is the radius of the circle. Solving for r, we have r = x / ( 2 p ) so the area of the circle is Acircle = p r2 = p [ x / ( 2 p ) ]2 = x2 / ( 4 p ). The 4 - x meters of wire remaining form the perimeter of a square, so that if we call the length of a side of this square s, we have 4 - x = P = 4s, and thus s = 1 - ( x / 4 ). Thus, the area of the square is Asquare = s2 = [ 1 - ( x / 4 ) ]2 = 1 - ( x / 2 ) + ( x2 / 16 ). Hence the combined area, which is what we want to maximize, is equal to

A = Acircle + Asquare = [ x2 / ( 4 p ) ] + [ 1 - ( x / 2 ) + ( x2 / 16 ) ].
As always, be find the critical numbers and then test critical points and the endpoints to locate our maximum (here, 0 < x < 4 is the domain of A). So consider
A ' (x) = [ ( 2x ) / ( 4 p ) ] + [ ( -1/2 ) + ( x / 8 ) ] = [ ( 1 / 2p ) + ( 1/8 ) ] x - ( 1/2 ),
which is always defined and is only equal to zero when x = ( 1/2 ) / [ ( 1 / 2p ) + ( 1/8 ) ] = 1.76. Now A '' (x) = [ ( 1 / 2p ) + ( 1/8 ) ] is always positive, so A '' (1.76) > 0. Thus, the area A(x) is minimum when x = 1.76 meters, yet we wanted to maximize the combined area. So we check the endpoints. When x = 0 (so all of the wire is used to construct a square) we have A(0) = 1 while when x = 4 (so all of the wire is used to make the circle) we have A(4) = 4 / p = 1.27, which is larger than the value at x = 0. Hence, the combined area is maximum when we invest all of the wire into making a circle of circumference 4 m and area 1.27 m2.

Problem 8:  A Norman window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass while the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. (Neglect the thickness of the frame.)

Solution:  Let h be the height of the rectangular part of the window and r be the radius of the semicircle on top, so the width of the window is the same as the diameter of the semicircle, which is 2r. The perimeter of the window is then the length of the semicirle's arc plus the height on each side plus the the base, so we have

P = pr + 2h + 2r
which is fixed. Then 2h = P - ( 2 + p ) r, so h = [ P - ( 2 + p ) r ] / 2. We want to admit as much light as possible for this perimeter P, while will be proportional to the area of the rectangle plus one-half the area of the semicircle, since the glass in the semicircle only admits half as much light as the clear glass in the rectangle. Hence, we want to maximize
L(r) = ( 2r ) h + ( 1/2 ) [ 1/2 p r2 ] = r [ P - ( 2 + p ) r ] + ( 1/4 ) p r2 = P r + [ -( 2 + p ) + ( p / 4 ) ] r2
So we find and test critical numbers:
L ' (r) = P + [ -( 2 + p ) + ( p / 4 ) ] [ 2r ],
which is defined for all r but is zero when r = P / [ 2 ( 2 + p ) - ( p / 2 ) ]. This is a maximum, as L '' (r) = -4 - 2p + ( p / 2 ) < 0. When r = P / [ 4 + 1.5p ], h = [ P - ( 2 + p ) ( P / [ 4 + 1.5p ) ] / 2 = P [ ( 4 + 1.5p ) - ( 2 + p ) ] / [ 2 ( 4 + 1.5p ]. So the proportions that maximize the amount of light transmitted by the window are for these values of r and h:
r = P / [ 4 + 1.5p ], h = P [ ( 2 + 0.5p ) / ( 4 + 1.5p ) ].

Problem 9:  We are going to cut a rectangular beam of width w and depth d out from a 12-in diameter cylindrical log.

  1. The strength S of a beam is proportional to its width times the square of its depth. Find the dimensions of the strongest beam that can be cut from a 12-in diameter cylindrical log.
  2. The stiffness T of a beam is proportional to its width times the cube of its depth. Find the dimensions of the stiffest beam that can be cut from a 12-in diameter cylindrical log.

Solution:  The cross-section of the cylindrical log will be a circle of diameter 12, in which we want to cut out a rectangle of width w and height d so maximize things, so we should cut out the rectangle so its four corners are on the boundary of the circle. Thus, if we draw the diagonal of the rectangle, it will be a diameter for the circle, so is of length 12, and the diagonal divides the rectangle into two right triangles whose legs have lengths w and d. Therefore, the Pythagorean Theorem says d2 + w2 = 122 = 144.

  1. To maximize the strength S = k w d2 = k w ( 144 - w2 ) = k ( 144 w - w3 ), find the critical number:
    S ' (w) = k ( 144 - 3 w2 )
    which is zero when w = [ 144 / 3 ]1/2 = 6.928, and this is a max since S '' (w) = k ( -6w ) = -6k w is negative whenever w > 0. Thus the strength of the beam will be maximized when we cut it from the log with width w = 6.928 in and depth d = [ 144 - ( 6.928 )2 ]1/2 = 9.798 in.
  2. To maximize the stiffness T = c w d3 = c w ( 144 - w2 )3/2, find its critical numbers:
    T ' (w) = ( cw ) [ ( 3/2 ) ( 144 - w2 )1/2 [ -2w ] ] + c ( 144 - w2 )3/2 = c ( 144 - w2 )1/2 [ ( -3w2 ) + ( 144 - w2 ) ] = c ( 144 - w2 )1/2 [ 144 - 4 w2 ]
    which is always defined for 0 < w < 12, but is zero when w = 12, and when w = [ 144 / 4 ]1/2 = 6. When w = 12, d = 0 so T = 0. When w = 0, T = 0 again, but when w = 6, d = 10.39 and T = 6734.21 c, so this is our maximum.

Problem 10:  The reaction of the body to a does of medicine can sometimes be represented by an equation of the form

R = M2 ( C/2 - M/3 ),
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, then R is measured in millimeters of mercury. If the reaction is a change in temperature, then R is measured in degrees, and so on. Find the amount of medicine to which the body is most sensitive by finding the value of M that maximizes the derivative dR/dM.

Solution:  What we want to maximize in this problem is not the function R, but rather its derivative dR/dM, which we will denote hereafter by S for "sensitivity". Thus S(M) = d/dM[ R ] = d/dM [ M2 ( C/2 - M/3 ) ] = M2 [ -1/3 ] + ( C/2 - M/3 ) [ 2M ] = -M2 / 3 + C M - M2 / 3 = C M - ( 2/3 ) M2. Thus consider

S ' (M) = C - ( 2/3 ) [ 2M ] = C - ( 4/3 ) M
which is always defined but is zero when M = ( 3/4 ) C. Moreover, S is maximized here, for S '' (M) = -4/3 for all M, so S '' (0.75 C) < 0. Hence the amount of medicine that maximizes the derivative dR/dM is M = 0.75 C.

November 8, 2004

Related Rates Worksheet from Nov. 8: problems and solutions

Problem 1  A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out string when the kite is 500 ft away from her?

Solution:  Consider the right triangle with the girl at one vertex, the kite at another, and the "shadow" of the kite vertically below it on the ground as the third, which is where we have the right angle. Suppose we call the length of string out at a given time S and the horizontal distance from the girl to the kite H. Then, by the Pythagorean Theorem, we have H2 + ( 300 )2 = S2. Then 2 H dH/dt + 0 = 2 S dS/dt, so dS/dt = ( H / S ) dH/dt. We know that the wind is blowing the kite horizontally away from the girl 25 ft/sec, so dH/dt = 25. When S = 500, using the Pythagorean Theorem, we find that H = 400, so the rate at which string must be let out at this moment is

dS/dt = [ ( 400 ) / ( 500 ) ] [ 25 ] = 20,
which means that the girl must let out 20 feet of string a second when the kite is 500 feet away from her.

Problem 2:  The length L of a rectangle is decreasing at a rate of 2 cm/sec while the width W is increasing at a rate of 2 cm/sec. When L = 12 cm and W = 5 cm, find the rates of change of the following:

  1. the area
  2. the perimeter
  3. the lengths of the diagonals of the rectangle
  4. Which of these quantities are increasing and which are decreasing?

Solution:  As the length L is decreasing 2 cm/sec, dL/dt = -2. Similarly, dW/dt = 2 since the width W is increasing at the rate 2 cm/sec.

  1. The area of the rectangle is A = LW, so dA/dt = ( L ) [ dW/dt ] + ( W ) [ dL/dt ]. Hence, when L = 12 and W = 5, we have
    dA/dt = ( 12 ) [ 2 ] + ( 5 ) [ -2 ] = 24 - 10 = 14,
    so the area of the rectangle is increasing 14 cm2/sec.
  2. The perimeter of the rectangle is P = 2L + 2W, so dP/dt = 2 [ dL/dt ] + 2 [ dW/dt ]. Thus
    dP/dt = 2 [ -2 ] + 2 [ 2 ] = -4 + 4 = 0
    implies that the perimeter is not changing, i.e., it is neither increasing nor decreasing or we may say that the perimeter is increasing at the rate of 0 cm/sec.
  3. The diagonals of the rectangle are related to the sides by the Pythagorean Theorem as D2 = L2 + W2, so 2 D dD/dt = 2 L dL/dt + 2 W dW/dt. Therefore, when L = 12 and W = 5, we have
    dD/dt = [ 2 ( 12 ) [ -2 ] + 2 ( 5 ) [ 2 ] ] / [ 2 ( 13 ) ] = [ -48 + 20 ] / [ 26 ] = -28/26 = -14/13 = -1.076923.
    Hence, the diagonals are decreasing at the rate of 1.077 cm/sec.
  4. As indicated in the solutions to the previous parts, the area is increasing, the perimeter is not changing, and the diagonals are decreasing.

Problem 3:  Sand falls from a conveyor belt at a rate of 10 m3/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. Find how fast

  1. the height
  2. the radius
of the pile are changing when the pile is 4 m high.

Solution:  We first set up the problem. Let h be the height of the pile of sand, D be the base diameter, and r the radius of the pile. Then we are told h = ( 3/8 ) D, and we know that D = 2r because the base is a circle of diameter D and radius r. Thus h = ( 3/8 ) [ 2r ] = ( 3/4 ) r and r = ( 4/3 ) h. We also interpret the fact that sand is being dumped onto the pile at the rate of 10 m3/min as stating that dV/dt = 10, where V will refer to the volume of sand in the pile. Finally, the volume V of the cone is related to its height h and radius r by the equation V = ( 1/3 ) [pi] r2 h.

  1. We are asked to find dh/dt when h = 4. The rate we do know is dV/dt = 10, so we want to relate the quantity h to the quantity V so that we may relate the unknown rate dh/dt to the known rate dV/dt = 10. We have the equation V = ( 1/3 ) [pi] r2 h, which is close but it unfortunately includes r still. However, we know that r = ( 4/3 ) h, so we substitute to find our relationships:
    V = ( 1/3 ) [pi] [ ( 4/3 ) h ]2 h = ( 16/27 ) [pi] h3,
    dV/dt = ( 16/27 ) [pi] [ 3 h2 dh/dt ].
    Thus, evaluating when h = 4 and dV/dt = 10, we have 10 = ( 16/27 ) [pi] [ 3 ( 4 )2 dh/dt ], so dh/dt = [ 270 / ( 768 [pi] ) ] = 0.1119. Therefore, the height of the pile is increasing at the rate of 0.1119 m/min when h = 4 m.
  2. To find dr/dt when h = 4, as we are asked to do in this next part, we could go back to the volume equation V = ( 1/3 ) [pi] r2 h and use h = ( 3/4 ) r to write V as a function of r, differentiate, and then evaluate. However, we also know that r = ( 4/3 ) h, which was our substitution in the first part, and we now know the value of dh/dt when h = 4, so let's use this simpler relation.
    dr/dt = ( 4/3 ) dh/dt
    implies that dr/dt = ( 4/3 ) [ 0.1119 ] = 0.1492, so the radius is growing at the rate of 0.1492 m/min when the height of the pile is 4 m.

Problem 4:  Water is flowing at a rate of 50 m3/min from a shallow concrete conical reservoir (vertex down) of base radius 45 m and height 6 m.

  1. How fast (cm/min) is the water level falling when the water is 5 m deep?
  2. How fast is the radius of the water's surface changing then? Again, answer in centimeters per minute.

Solution:  As above, we set up the problem by naming the depth of water in the reservoir h and the radius of the water's surface r. Then the right triangle with legs h and r is similar to the right triangle for the cone of legs 6 and 45, respectively. Thus, h/r = 6/45, so h = ( 6/45 ) r and r = ( 45/6 ) h. Once more, the volume of water in the cone is V = ( 1/3 ) [pi] r2 h and we are told that water is flowing out of the reservoir at a rate of 50 m3/min, which means that dV/dt = -50 (because the volume is decreasing).

  1. We want to know how fast the water level is falling (so this already takes into account that dh/dt should be negative and we will give the absolute value of dh/dt as our answer) when h = 5. As we do know that dV/dt = -50 and V = ( 1/3 ) [pi] r2 h and r = ( 15/2 ) h, we have
    V = ( 1/3 ) [pi] [ ( 15/2 ) h ]2 h = ( 225/12 ) [pi] h3
    dV/dt = ( 225/12 ) [pi] [ 3 h2 dh/dt ]
    Evaluating this when h = 5 given that dV/dt = -50, we have -50 = ( 225/12 ) [pi] [ 3 ( 5 )2 dh/dt ] which implies that dh/dt = ( -600 ) / ( 16875 [pi] ) = -0.0113176848. Therefore, the water level in the reservoir is falling at a rate of 0.0113 m/min = 1.13 cm/min.
  2. As we found that r = ( 45/6 ) h, we have dr/dt = ( 45/6 ) dh/dt. When h = 5, we know that dh/dt = -1.13 cm/min, so dr/dt = ( 45/6 ) [ -1.13 ] = -8.475 cm/min and we conclude that the radius of the water's surface is decreasing at a rate of 8.475 cm/min when the water level is 5 m.

Problem 5:  A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft/sec.

  1. How fast is the boat approaching the dock when 10 ft of rope are out?
  2. At what rate is the angle between the rope and the dock changing then?

Solution:  Let R be the length of the rope from the bow of the boat to the dock and H be the horizontal distance from the boat to the dock.

  1. By the Pythagorean Theorem, we have R2 = H2 + 36, so 2 R dR/dt = 2 H dH/dt. Also using the Pythagorean Theorem, when R = 10, H = 8, and we know that dR/dt = -2 since the rope is being hauled in at the rate of 2 ft/sec. Hence
    dH/dt = ( R / H ) dR/dt = [ ( 10 ) / ( 8 ) ] [ -2 ] = -2.5
    which means that the boat is approaching the dock at a rate of 2.5 ft/sec (because "approaching" takes into account that the distance between the boat and the dock is decreasing).
  2. Let A denote the angle between the rope and the dock, so tan A = H / 6. Thus [ sec2 A ] dA/dt = ( 1/6 ) dH/dt. We know that dH/dt = -2.5 and we know that cos A = adj/hyp = ( 6 ) / ( 10 ) when R = 10, so
    dA/dt = [ cos2 A ] ( 1/6 ) [ dH/dt ] = [ 6/10 ]2 ( 1/6 ) [ -2.5 ] = -0.15
    which says that the angle is changing at the rate of -0.15 radians/sec when R = 10.

Problem 6:  A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate 5 ft/sec.

  1. How fast is the top of the ladder sliding down the wall then?
  2. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
  3. At what rate is the angle between the ladder and the ground changing then?

Solution:  Let G be the distance between the base of the ladder and the wall and W be the height of the ladder on the wall. Then G2 + W2 = 132 and we are told that when G = 12, dG/dt = 5.

  1. Differentiating G2 + W2 = 169 with respect to time t, we have 2 G dG/dt + 2 W dW/dt = 0. Substituting G = 12, W = 5, and dG/dt = 5, we find
    dW/dt = ( -G / W ) dG/dt = [ ( -12 ) / ( 5 ) ] [ 5 ] = -12,
    i.e., the ladder is sliding down the wall 12 ft/sec.
  2. The area of the triangle is A = G W / 2, so dA/dt = 1/2 [ G dW/dt + W dG/dt ]. Therefore,
    dA/dt = ( 1/2 ) [ ( 12 ) [ -12 ] + ( 5 ) [ 5 ] ] = -59.5
    so the area is decreasing 59.5 ft2/sec.
  3. Let a be the angle between the ladder and the ground. Then tan a = W / 13 = ( 1/13 ) W, so [ sec2 a ] da/dt = ( 1/13 ) dW/dt. When G = 12 and W = 5, sec a = hyp/adj = 13/12, so
    da/dt = [ cos a ]2 ( 1/13 ) dW/dt = [ 12/13 ]2 ( 1/13 ) [ -12 ] = -0.7865
    and the angle is decreasing at the rate of 0.7865 radians/sec.

Problem 7:  A baseball diamond is a square 90 ft on a side.

  1. A player runs from first base to second at a rate of 16 ft/sec. At what rate is the player's distance from third base changing when the runner is 30 ft from first base?
  2. Now suppose the runner slides into second at a rate of 15 ft/sec. At what rate is his distance from third base changing as the player touches base?

Solution:  Let x be the distance the runner has run so far from first to second base, so dx/dt is the runner's velocity. The distance the runner still has to run is ( 90 - x ) feet and this distance together with the 90 feet from 2nd to 3rd base form two legs of a right triangle. The hypothenuse of the right triangle, which we'll call y, is the distance from the runner to third base. The Pythagorean Theorem relates y to x by y2 = 902 + ( 90 - x )2. In both parts of the problem, we want to find how the distance y changes, i.e., what is dy/dt.

  1. When x = 30, we are told that dx/dt = 16. Differentiating y2 = 902 + ( 90 - x )2 with respect to time, we find 2 y dy/dt = 2 ( 90 - x ) [ -dx/dt ], so
    dy/dt = [ ( 90 - x ) / y ] [ -dx/dt ].
    Evaluating this when x = 30 (so y = [ ( 90 )2 + ( 60 )2 ]1/2 = 108.1665 ft) and dx/dt = 16, we have dy/dt = [ ( 60 ) / ( 108.1665 ) ] [ -16 ] = -8.875, which means that the distance from the runner to third base is decreasing at the rate of 8.875 ft/sec at this time.
  2. For this part of the problem, we again use the relationship dy/dt = [ ( 90 - x ) / y ] [ -dx/dt ], but now with x = 90 (because we're interested in the moment when the runner reaches second base, so he has run the full 90 feet from first base) and dx/dt = 15. Note that when x = 90, y = 90, so dy/dt = [ ( 0 ) / ( 90 ) ] [ -15 ] = 0. That is, at the moment the runner touches second base, the rate of change of his distance from third base is zero (it isn't changing at that moment). This makes sense as up to the point of reaching second, the distance y has been decreasing but if the runner were to continue running past second, the distance y would begin to increase as soon as he passes second. Thus y changes from decreasing to increasing when x = 90, so dy/dt changes from being negative to positive when x = 90, so it is reasonable that dy/dt = 0 when x = 90.

Problem 8:  Consider the curve x2 + xy + y2 = 7.

  1. Find the points where the tangent is parallel to the x-axis.
  2. Find the points where the tangent is parallel to the y-axis.
  3. Find the two points where the curve crosses the x-axis and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents?

Solution:  We first implicitly differentiate the equation to obtain 2x + x dy/dx + y + 2 y dy/dt = 0, so ( x + 2y ) dy/dx = -2x - y and dy/dx = - ( 2x + y ) / ( x + 2y ).

  1. The curve is parallel to the x-axis when the tangent line is horizontal, which happens when dy/dx = 0, i.e., when 2x + y = 0 on the curve. This occurs where y = -2x and x2 + xy + y2 = 7, which we solve by substitution: x2 + x ( -2x ) + ( -2x )2 = x2 - 2x2 + 4x2 = 3x2, so x2 = 7/3 and x = +( 7/3 )1/2 = +1.5275. At x = +1.5275, we have from the required equation y = -2x that the point on the curve we're after is (+1.5275,-3.055), while the point associated to x = -1.5275 is (-1.5275,3.055). Thus the points on the curve x2 + xy + y2 = 7 where the tangent line is paralle to the x-axis are (1.5275,-3.055) and (-1.5275,3.055).
  2. The curve is parallel to the y-axis when the tangent line is vertical, which happens when dy/dx is not defined, i.e., when x + 2y = 0. Thus we want to solve the pair of equations x2 + xy + y2 = 7 and x = -2y, which we do by substitution: ( -2y )2 + ( -2y ) y + y2 = 3y2 = 7 when y = +1.5275 and x = -2y implies the tangent line to the curve is vertical at the points (-3.055,1.5272) and (3.055,-1.5275).
  3. For this last part, we need to find the points where the curve crosses the x-axis, which happens when y = 0 so that x2 = 7, i.e., x = +2.64575. At these points, the tangents to the curve have slope:
    m(2.64575,0) = - [ 2 ( 2.64575 ) + ( 0 ) ] / [ ( 2.64575 ) + 2 ( 0 ) ] = - 2
    m(-2.64575,0) = - [ 2 ( -2.64575 ) + ( 0 ) ] / [ ( -2.64575 ) + 2 ( 0 ) ] = - 2
    which are the same so the tangent lines are parallel and both of slope m = -2.

Problem 9:  Use the technique of logarithmic differentiation to find dy/dx.

  1. y = xln x
  2. y = x( 1 / ln x )
  3. y = [ x ( x2 + 1 )1/2 ] / [ ( x + 1 )2/3 ]

Solution

  1. Setting y = xln x, take the natural log of both sides to get ln y = ln [ xln x ] = ( ln x ) ln [ x ] = ( ln x )2. Thus ( 1 / y ) dy/dx = 2 ( ln x ) [ 1 / x ], so
    dy/dx = y [ 2 ( ln x ) / x ] = [ xln x ] [ ( 2 ln x ) / x ] = 2 ( ln x ) x( ln x ) - 1.
  2. For y = x( 1 / ln x ), taking ln of both sides gives us ln y = ln [ x( 1 / ln x ) ] = ( 1 / ln x ) ln [ x ] = 1, so ( 1 / y ) dy/dx = 0. Therefore, dy/dx = 0.
  3. Taking ln, we have ln y = ln x + ( 1/2 ) ln( x2 + 1 ) - ( 2/3 ) ln( x + 1 ), so ( 1 / y ) dy/dx = 1/x + ( 1/2 ) [ 1 / ( x2 + 1 ) ] [ 2x ] - ( 2/3 ) [ 1 / ( x + 1 ) ] [ 1 ] = 1/x + [ x / ( x2 + 1 ) ] - [ 2 / 3( x + 1 ) ]. Hence
    dy/dx = y [ 1/x + ( x / ( x2 + 1 ) ) - ( 2 / 3( x + 1 ) ) ] = [ x ( x2 + 1 )1/2 ] / [ ( x + 1 )2/3 ] [ 1/x + ( x / ( x2 + 1 ) ) - ( 2 / 3( x + 1 ) ) ].

Problem 10:  Consider the following four functions: f(x) = ex - 1, g(x) = x3 + x, h(x) = ln( x + 1 ), j(x) = sin x.

  1. Find the linearizations of f, g, h and j near x = 0.
  2. Given your answer to part (a), what is special about those four particular functions that made the result unusual?
  3. For each function, note how long it takes before the approximation fails badly. (For the purposes of discussion, we can define "fails badly" as deviating from its actual value by more than 1/2.) For which function is the approximation best? For which is it worst? Explain why this is so.

Solution:  Suppose F(x) is a function. The linearization of F(x) near x = a is the function L(x) = F(a) + F'(a) [ x - a ], which is the equation of the tangent line to the graph y = F(x) at the point x = a.

  1. The linearization of f(x) = ex - 1 is therefore
    Lf(x) = f(0) + f'(0) x = [ 0 ] + [ e0 ] x = x
    because f'(x) = ex. The linearization of g(x) = x3 + x is
    Lg(x) = g(0) + g'(0) x = [ 0 ] + [ 3 ( 0 )2 + 1 ] x = x
    because g'(x) = 3x2 + 1. The linearization of h(x) = ln( x + 1 ) is
    Lh(x) = h(0) + h'(0) x = [ 0 ] + [ 1 / ( ( 0 ) + 1 ) ] x = x
    because h'(x) = 1 / ( x + 1 ). The linearization of j(x) = sin x is
    Lj(x) = j(0) + j'(0) x = [ 0 ] + [ cos 0 ] x = x
    since j'(x) = cos x.
  2. All of these functions have the same linearizations near x = 0 because they all have the same value at x = 0 (namely 0) and their slopes are all the same at x = 0 (namely 1). Therefore, all of the graphs of these four functions have exactly the same tangent line at x = 0, which is why their linearizations are the same.
  3. The linear approximation Lf is within the allowed error of +0.5 for x roughly between -1.2 and 0.85 (these values were found using a graphing calculator). The linearization Lg is within the allowed error of +0.5 for x roughly between -0.8 and 0.8. The linearization Lh is within the allowed error of +0.5 for x roughly between -0.7 and 1.4. Finally, the linearization Lj is within the allowed error of +0.5 for x roughly between -1.5 and 1.5. Thus, for this tolerance of +0.5, the linear approximation L(x) = x is "best" for j(x) = sin x since it has the largest interval on which the two are within 0.5 of one another (in fact, all of the other intervals are contained in this one for the sin x) and it is "worst" for g(x) = x3 + x since the interval for this one is the smallest (and is very nearly contained in the intervals for all of the others).

Problem 11:  Find a linear approximation for y = x-1/2 at x = 4, and use differentials to approximate dy for dx = 1 and dx = -1.

Solution:  L(x) = f(4) + f'(4) [ x - 4 ], so we need to find out the value of f(4) = ( 4 )-1/2 = 1/2 and f'(4). Now f'(x) = ( -1/2 ) x-3/2, so f'(4) = ( -1/2 ) ( 4 )-3/2 = -1/16. Thus the linear approximation for y = x-1/2 at x = 4 is

L(x) = ( 1/2 ) + ( -1/16 ) [ x - 4 ].
For the second part of the problem, using differentials to approximate dy when dx = +1 and x = 4, we recall that dy = f'(x) dx = [ ( -1/2 ) x-3/2 ] dx. Hence, when dx = 1,
dy = [ ( -1/2 ) ( 4 )-3/2 ] [ 1 ] = -1/16,
and, when dx = -1,
dy = [ ( -1/2 ) ( 4 )-3/2 ] [ -1 ] = 1/16.

Problem 12:  A sphere with radius 100 cm is to be painted with a layer of paint 0.02 cm thick. Use differentials to estimate the volume of paint required.

Solution:  The volume of paint required to paint this sphere is the change in volume of the sphere when its radius is increased from 100 (the radius of the sphere) to 100.02 (the radius of the sphere plus paint). However, using differentials, we can approximate this using the formula V = ( 4/3 ) [pi] r3 to find dV = ( 4/3 ) [pi] [ 3 r2 dr ]. Thus the approximate volume of paint required is the value of dV when r = 100 and dr = 0.02, i.e.,

dV = ( 4/3 ) [pi] [ 3 ( 100 )2 ( 0.02 ) ] = 2513.274.
Therefore, we estimate that 2513.274 cm3 of paint is needed.
November 5, 2004

Problem 1:  Suppose the earth were a perfect sphere and we determined its radius to be 3959 + 0.1 miles. What effect does the tolerance +0.1 have on our estimate of the surface area of the earth?

Solution:  The formula for the surface area of a sphere in terms of its radius is S = 4 [pi] r2, so the corresponding differentials are dS = 8 [pi] r dr. Thus, with the tolerance dr = 0.1 and r = 3959, our estimate of the surface area of the earth may be off by as much as dS = 8 [pi] ( 3959 ) ( 0.1 ) = 9950 square miles, which is roughly the size of the state of Maryland.

Problem 2:  About how accurately should we measure the radius of a sphere to calculate its volume V = ( 4/3 ) [pi] r3 within 1%?

Solution:  We want dS = 8 [pi] r dr to be less than or equal to S/100 = [ 4 [pi] r2 ] / 100. Thus we want 8 [pi] r dr < [ 4 [pi] r2 ] / 100, so dr < [ 4 [pi] r2 ] / [ 100 * ( 8 [pi] r ) ] = r / 200 = ( 1/2 ) ( r/100 ). Thus, we need to measure r within 0.5% of its actual value to ensure that our calculation of S is within 1% of its real value.

Problem 3:  (Clogged arteries and angioplasty) In the late 1830's, French physiologist Jean Poiseuille discovered the formula used today to predict how much the radius of a clogged artery has to be expanded to restore normal blood flow. His formula,

V = k r4

says the volume V of fluid flowing through a small tube or pipe in a unit of time at a fixed pressure is a constant times the 4th power of the radius. Based on this, how will a 10% increase in r affect V?

Solution:  Given V = k r4, we have dV = k [ 4 r3 dr ] = 4k r3 dr. Then dV / V = [ 4k r3 dr ] / [ k r4 ] = [ 4 dr ] / [ r ] = 4 [ dr / r ]. Therefore, a 10% increase in r will produce a 4 [ 10% ] = 40% increase in blood flow.

Problem 4:  The normal to a curve y = f(x) at the point x = a is the line through the point (a,f(a)) perpendicular to the tangent line there. Thus, the slope of the normal is the opposite reciprocal of the slope of the tangent line, so m = -1 / f'(a).

  1. Find equations for the tangent and normal to the cissoid of Diocles y2 ( 2 - x ) = x3 at (1,1).
  2. The line that is normal to the curve x2 + 2xy - 3y2 = 0 at (1,1) intersects the curve at what other point?
  3. Show that if it is possible to draw three normals from the point (a,0) to the parabola x = y2, then a must be greater than 1/2. One of the normals is the x-axis. For what value of a are the other two normals perpendicular?

Solution

  1. To begin with, we want to find the value of dy/dx at the point (1,1) on the cissoid of Diocles, y2 ( 2 - x ) = x3. Implicitly differentiating, we obtain y2 [ -1 ] + ( 2 - x ) [ 2y dy/dx ] = 3x2. Solving for dy/dx, we have 2y ( 2 - x ) dy/dx = 3x2 + y2, so dy/dx = [ 3x2 + y2 ] / [ 2y ( 2 - x ) ]. Thus, at (1,1), the tangent line to the cissoid has slope
    dy/dx = [ 3 (1)2 + (1)2 ] / [ 2 ( 1 ) ( 2 - ( 1 ) ) ] = [ 3 + 1 ] / [ 2 ] = 2.
    Hence the equation of the tangent line is
    y - 1 = 2 ( x - 1 ).
    The normal is the line perpendicular to the tangent line also passing through the point (1,1), so its slope is -1 / 2 and its equation is
    y - 1 = ( -1/2 ) ( x - 1 ).
  2. Find the equation of the normal to x2 + 2xy - 3y2 = 0 by finding the slope of the tangent line to which it is perpendicular. This slope is the value of dy/dx at (1,1), and we find dy/dx implicitly: 2x + ( 2x ) [ dy/dx ] + ( y ) [ 2 ] - 3 [ 2y dy/dx ] = 0 implies that 2x + 2y = 6y dy/dx - 2x dy/dx = ( 6y - 2x ) dy/dx and thus that dy/dx = ( 2x + 2y ) / ( 6y - 2x ). Hence the slope of the tangent line at (1,1) is dy/dx = [ 2 ( 1 ) + 2 ( 1 ) ] / [ 6 ( 1 ) - 2 ( 1 ) ] = 4 / 4 = 1. This means the slope of the normal at (1,1) is m = -1/1 = -1. Thus the equation of the normal is
    y - 1 = -1 ( x - 1 ).
    To find the other point on the curve x2 + 2xy - 3y2 = 0 where the normal y - 1 = -x + 1 (or y = -x + 2) intersects the curve, substitute y = 2 - x into the equation
    x2 + 2xy - 3y2 = x2 + 2x( 2 - x ) - 3 ( 2 - x )2 = 0
    and solve for x:
    x2 + 4x - 2x2 - 3 ( 4 - 4x + x2 ) = -12 + 16x - 4x2 = -4 ( x2 - 4x + 3 ) = -4 ( x - 1 ) ( x - 3 ) = 0
    when x = 1 or x = 3. When x = 1, from the normal line we get y = 1 and recover our first point of intersection, (1,1), again. Using x = 3, we find that y = 2 - ( 3 ) = -1, so the point ( 3,-1 ) is the second point of intersection of the normal line y = 2 - x with the curve x2 + 2xy - 3y2 = 0.
  3. Let's first find what the slope of the tangent line to the curve x = y2 is at a point (a,b). Thus we need to find the derivative dy/dx implicitly:
    1 = 2y dy/dx implies that dy/dx = 1 / [ 2y ].
    So, at the point with coordinates (y2,y) on x = y2, the tangent line has slope 1 / [ 2y ]. At that same point, the line joining the point (a,0) to (y2,y) has slope
    m = [ y - 0 ] / [ y2 - a ] = y / [ y2 - a ].
    For these lines to be perpendicular, and thus for the segment joining (a,0) to (y2,y) to be normal to x = y2, the product of the slope m = y / [ y2 - a ] with the tangent line slope 1 / [ 2y ] must be equal to -1:
    -1 = [ 1 / ( 2y ) ] [ y / ( y2 - a ) ] = 1 / [ 2 ( y2 - a ) ],
    so y2 - a = -1/2. Thus
    a = 1/2 + y2 > 1/2
    for all y. Thus, for there to be three points where the normals to x = y2 pass through (a,0), for two of the points we must have y different from 0 and thus we must have a = 1/2 + y2 strictly larger than 1/2 as claimed. To find the value of a where the other two lines are perpendicular, we need to find a so that both
    a = 1/2 + y2 and [ y / ( y2 - a ) ] [ -y / ( y2 - a ) ] = -1
    (the first requirement is to ensure that the segments joining (a,0) to (y2,+y) are normal to x = y2 and the second to ensure that these segments are perpendicular to each other). The second equation becomes y2 = ( y2 - a )2 = y4 - 2ay2 + a2 upon simplifying and clearing denominators. If we now use the first, in the form y2 = a - 1/2, to replace y2 and y4 we get
    y2 = y4 - 2ay2 + a2
    a - 1/2 = ( a - 1/2 )2 - 2a ( a - 1/2 ) + a2 = ( a2 - a + 1/4 ) - 2a2 + a + a2 = 1/4.
    Thus a - 1/2 = 1/4, so a = 3/4 is our solution.

Problem 5:  (The melting ice cube) How long does it take an ice cube to melt if its volume decreases 25% in the first hour? (Assume the ice cube is actually a cube, so its volume V is equal to s3, where s is the length of any of its sides. The rate at which the volume changes with respect to time is proportional to its surface area (as this is where the melting occurs), so dV/dt = -k S for some positive constant k, where S is the surface area.)

Solution:  First of all, if our ice cube is literally a cube of side length s, then its volume is V = s3 and its surface area is S = 6s2, since it has six sides each of area s2. Thus, we are given that dV/dt = -k S = -k ( 6s2 ), but we can also calculate dV/dt = [ dV/ds ] [ ds/dt ] = [ 3 s2 ] ds/dt by the chain rule from the formula V = s3. Thus 3s2 ds/dt = -k ( 6s2 ), so ds/dt = -2k. That is, the side length is decreasing at a constant rate of 2k units per hour. Thus, if s0 is the the initial side length of the ice cube, the length of its side one hour later is s1 = s0 - 2k. Therefore

2k = s0 - s1.
The melting time is the value of t that makes 2kt = s0 (because the side decreases by 2k units per hour, so after t hours the length has been reduced by 2kt and we want to know at what time has all of the length vanished). Hence,
tmelt = s0 / [ 2k ] = [ s0 ] / [ s0 - s1 ] = [ 1 ] / [ 1 - ( s1 / s0 ) ].
But
s1 / s0 = [ ( 3/4 ) V ]1/3 / [ V1/3 ] = ( 3/4 )1/3 = 0.91
Therefore,
tmelt = 1 / [ 1 - 0.91 ] = 11 hours.

Problem 6:  (Temperature and the period of a pendulum) For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation

T = 2 [pi] ( L / g )1/2

where g is the constant acceleration of gravity at the pendulum's location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,
dL/du = kL.

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT/2.

Solution:  The rate of change of the period T with respect to the temperature u is the derivative dT/du. However, T is given as a function of the length L and we aren't told how to express L as a function of u directly, but only that dL/du = kL. Still, by the chain rule, this should be enough, for

dT/du = [ dT/dL ] [ dL/du ]
and dT/dL = d/dL [ 2 [pi] ( L / g )1/2 ] = ( 2 [pi] ) [ 1/2 ( L / g )-1/2 [ 1 / g ] ] = [pi] / [ g ( L / g )1/2 ] = [pi] / [ ( gL )1/2 ]. Hence
dT/du = [ dT/dL ] [ dL/du ] = [ [pi] / ( gL )1/2 ] [ kL ] = [ k [pi] L ] / [ ( gL )1/2 ] = k [pi] ( L / g )1/2 = kT/2.

Tangent Lines and Rates of Change

Problem 1:  Suppose that the distance an aircraft travels along a runway before takeoff is given by D = ( 10 / 9 ) t2, where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. If the aircraft will become airborne when its speed reaches 200 km/hr, how long must the runway be for this aircraft?

Solution:  The length of the runway must be long enough for the aircraft to have enough time in order for its speed to reach 200 km/hr. The speed of the aircraft is the absolute value of its velocity and the velocity is the derivative of its position function D = ( 10/9 ) t2. Thus v(t) = dD/dt = d/dt [ ( 10/9 ) t2 ] = ( 10/9 ) [ 2t ] = ( 20/9 ) t. However, the distances for D were measured in meters and the time was measured in seconds, but the speed we want is expressed in km/hr. Therefore, we need to convert the speed 200 km/hr into m/sec by recalling that there are 1000 m per km and 3600 seconds per hour, so the speed of the aircraft must be 55.555... m/sec to become airborne. Thus we want to know when v(t) = 200,000, i.e., when (20/9) t = 55.55555. Dividing both sides of this equation by 20/9, we obtain t = 25 sec. Thus, to decide how long the runway must be for this aircraft, we determine how far it travels in 25 sec, i.e., D(25) = ( 10/9 ) (25)2 = 694.444 meters.

Problem 2:  Suppose it costs C(x) = x3 - 6x2 + 15x dollars to produce x radiators when 8 to 30 radiators are produced per day. You currently produce 10 radiators. About how much extra will it cost to produce one more a day?

Solution:  To estimate the additional cost of producing one more radiator, we find the marginal cost function's value when x = 10. The marginal cost is C ' (x) = d/dx [ x3 - 6x2 + 15x ] = 3x2 - 12x + 15, so C ' (10) = 3(10)2 - 12(10) + 15 = 300 - 120 + 15 = 195. Thus the additional cost of producing the 11th radiator will be approximately $195.

Problem 3:  When a bactericide was added to a nutrient broth in which bacteria were growing, the bacteria population continued to grow for awhile, then stopped growing and then declined. The size of the population at time t (hours) was b = 106 + 104 t - 103 t2. Find the growth rates at times (a) t = 0, (b) t = 5, and (c) t = 10.

Solution:  The growth rates of the bacteria culture are given by the derivative db/dt, which is the rate of change of the population with respect to time. So, we must first determine the function db/dt = d/dt [ 106 + 104 t - 103 t2 ] = 104 - 103 [ 2t ]. Thus, the growth rates are

Problem 4:  Define f(1) in a way that extends f(s) = [ s3 - 1 ] / [ s2 - 1 ] to be continuous at s = 1.

Solution:  In order for f to be continuous at s = 1, it must satisfy three things

Yet f(1) is not defined since when s = 1 the denominator s2 - 1 in the definition of f is zero. Still, all is not lost, for if the limit as s approaches 1 does exist, then we can define f(1) to equal this limit and thus all three requirements above would then be satisfied and the extended function will be continuous at s = 1. So define f(1) to be equal to lims --> 1 [ s3 - 1 ] / [ s2 - 1 ] = lims --> 1 [ ( s - 1 )( s2 + s + 1 ) ] / [ ( s - 1 )( s + 1 ) ] = lims --> 1 [ s2 + s + 1 ] / [ s + 1 ] = [ (1)2 + (1) + 1 ] / [ (1) + 1 ] = 3/2. Hence, by defining f(1) = 3/2, the extended function is continuous at s = 1.

Problem 5:  For what value of A is the function f(x) defined by f(x) = x2 - 1 for x < 3 and by f(x) = 2Ax for x > 3 continuous at every x? Is the function differentiable for this value of A?

Solution:  No matter what value of A we select, the functions x2 - 1 and 2Ax are continuous everywhere, so the only problem can occur where they "meet" at x = 3, i.e., when f switches from being defined as x2 -1 to being 2Ax. Recall that continuity requires that f is defined at x = 3, which it is: f(3) = 2A(3) = 6A, by definition of the function f above. Now we need the limit, as x approaches 3, of f(x) to exist, which will only happen if the right-hand limit and left-hand limit both exist and are equal. Both of the one-sided limits will exist since each of the functions x2 - 1 and 2Ax were continuous, and the respective limits are the values of these two functions at x = 3. So the left-hand limit is (3)2 - 1 = 9 - 1 = 8 and the right-hand limit is 2A(3) = 6A. For the limit of f to exist at 3, then, we need to have 8 = 6A or A = 8/6 = 4/3. Therefore, if A = 4/3, then f(3) is defined, the limit as x approaches 3 of f(x) exists, and this limit is equal to the value of f(3) (both are equal to 4). The second question asks whether the function, with A = 4/3, is differentiable. Again, as the functions x2 - 1 and 2(4/3)x = (8/3)x are each differentiable everywhere, the function f is differentiable everywhere with the only possible problem, again, can arise where f changes from being given by x2 - 1 to being (8/3)x at x = 3. For f to be differentiable here, we would need the left-hand and right-hand derivatives (i.e., the derivatives of x2 - 1 and (8/3)x at x = 3) to be equal. Yet the left-hand derivative is the value of d/dx[ x2 - 1 ] = 2x at x = 3, so it is 2(3) = 6. The right-hand derivative is equal to the value of d/dx [ (8/3) x ] = 8/3 at x = 3, which is 8/3. As these two are not equal, the function f is not differentiable at x = 3, and therefore f is not differentiable.

Problem 6:  Give an example of functions f and g, both continuous at x = 0, for which the composition f o g is not continuous at x = 0.

Solution:  The thing that I must stress first is that there is not just one right answer to this problem, since there are several examples that I could think of at the moment. But here is one possible example. Let f(x) = 1 / ( 1 + x ) and g(x) = x - 1. Both are continuous at x = 0 because f is a rational function with x = 0 in its domain and g is a polynomial, so it is continous everywhere. However, f(g(x)) = f( x - 1 ) = 1 / [ 1 + ( x - 1 ) ] = 1 / x is not continuous at x = 0.