Fall Term, 2004

**Instructor**:
David 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).

**Problem 1**: A poster is to have an area of 180 in^{2}
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

**Problem 2**: Your iron works has contracted to design and build
a 500 ft^{3}, 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 = x^{2} y, which must equal 500 according to the contract. Thus
y = 500 / x^{2} = 500 x^{-2}. Again, we want to minimize
the surface area of the *open-top* rectangular tank, which is

**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 S_{1}
and S_{2} = 3 S_{1} 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

**Problem 4**: Let v_{1} be the velocity of light in air
and v_{2} 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

**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 [ h^{2} + x^{2} ]^{1/2}
while the distance from C to B is [ d^{2} + ( D - x )^{2}
]^{1/2}. As light travels at velocity v_{1} in the air (i.e.,
on its path from A to C) and v_{2} in the water (i.e., from C to B),
the time it takes light to travel from A to C is

**Problem 5**: You have been asked to design 1000-cm^{3}
cans shaped like right circular cylinders.

- 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 r ^{2}+ 2 p r h. - 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 r rather than S = 2 p r^{2}+ 2 p r h^{2}+ 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 cm^{3}, and the volume of a
cylinder is p r^{2} h. Hence
p r^{2} h = 1000, so h = 1000 /
( p r^{2} ) and we'll write S and A in terms
of r only using this substitution and the domains of each will be { r > 0 }.

- We are asked to minimize S = 2 p r
^{2}+ 2 p r h = 2 p r^{2}+ 2 p r ( 1000 / p r^{2}) = 2 p r^{2}+ 2000 r^{-1}. Therefore, as always, let's find the critical numbers:S ' (r) = 4 p r - 2000 r which is zero when r = ( 500 / p )^{-2}= 4 [ p r^{3}- 500 ] / r^{2},^{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. - Now we want to minimize the function A(r) = 8 r
^{2}+ 2 p r [ 1000 / ( p r^{2}) ] = 8 r^{2}+ 2000 r^{-1}. As above, considerA ' (r) = 16 r - 2000 r which is defined everywhere on its domain, { r > 0 }, but is zero when r^{-2}= 16 [ r^{3}- 125 ] / r^{2},^{3}= 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.

**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 = x^{2} 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 = x^{2} 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.

- If 0
__<__x__<__y, then the third dimension y is the length and the girth is 4x, so the constraint is y + 4x = 108, so that y = 108 - 4x. Thus we want to maximize V = x^{2}y = x^{2}( 108 - 4x ) = 108 x^{2}- 4 x^{3}. Thus, consider V ' (x) = 216 x - 12 x^{2}= 12 x ( 18 - x ), which is always defined and is only zero when x = 0 or x = 18. When x = 0, V = 0 is minimum, which isn't what we wanted to do. When x = 18, V '' (x) = 216 - 24 x is equal to V '' (18) = -216 < 0, so that volume is maximized when x = 18 and y = 108 - 4(18) = 36, and this maximum volume is V(18) = ( 18 )^{2}( 36 ) = 11664 in^{3}. - Now, if 0
__<__y < x, so that one of the sides of the square base is the longest side, then the U.S. Postal Service's constraint becomes x + ( 2x + 2y ) = 3x + 2y = 108, so that y = 54 - 1.5 x. Then V = x^{2}( 54 - 1.5 x ) = 54 x^{2}- 1.5 x^{3}, so V ' (x) = 108 x - 4.5 x^{2}= 9 x ( 12 - 0.5 x ), which is always defined and is zero either when x = 0 (can't happen, since x > y__>__0) or x = 24 in which case y = 18 and V = 10368 in^{3}. This critical number x = 24 does correspond to a local max for V, as V '' (x) = 108 - 9 x is negative when x = 24 (V '' (24) = -108 < 0).

**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 A_{circle} = p
r^{2} = p [ x / ( 2 p
) ]^{2} = x^{2} / ( 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 A_{square}
= s^{2} = [ 1 - ( x / 4 ) ]^{2} = 1 - ( x / 2 ) + (
x^{2} / 16 ). Hence the combined area, which is what we want to
*maximize*, is equal to

**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

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

- 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.
- 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 d^{2} + w^{2} = 12^{2} = 144.

- To maximize the strength S = k w d
^{2}= k w ( 144 - w^{2}) = k ( 144 w - w^{3}), find the critical number:S ' (w) = k ( 144 - 3 w which is zero when w = [ 144 / 3 ]^{2})^{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. - To maximize the stiffness T = c w d
^{3}= c w ( 144 - w^{2})^{3/2}, find its critical numbers:T ' (w) = ( cw ) [ ( 3/2 ) ( 144 - w which is always defined for 0^{2})^{1/2}[ -2w ] ] + c ( 144 - w^{2})^{3/2}= c ( 144 - w^{2})^{1/2}[ ( -3w^{2}) + ( 144 - w^{2}) ] = c ( 144 - w^{2})^{1/2}[ 144 - 4 w^{2}]__<__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

**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 [
M^{2} ( C/2 - M/3 ) ] = M^{2} [ -1/3 ] + ( C/2 - M/3 )
[ 2M ] = -M^{2} / 3 + C M - M^{2} / 3 = C M - ( 2/3 )
M^{2}. Thus consider

**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 H^{2} + ( 300 )^{2} = S^{2}. 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

**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:

- the area
- the perimeter
- the lengths of the diagonals of the rectangle
- 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.

- 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 cm^{2}/sec. - 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. - The diagonals of the rectangle are related to the sides by the Pythagorean
Theorem as D
^{2}= L^{2}+ W^{2}, so 2 D dD/dt = 2 L dL/dt + 2 W dW/dt. Therefore, when L = 12 and W = 5, we havedD/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. - 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 m^{3}/min onto the top of a conical pile. The height of the pile
is always three-eighths of the base diameter. Find how fast

- the height
- the radius

**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 m^{3}/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] r^{2} h.

- 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] r
^{2}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 ] Thus, evaluating when h = 4 and dV/dt = 10, we have 10 = ( 16/27 ) [pi] [ 3 ( 4 )^{2}h = ( 16/27 ) [pi] h^{3},

dV/dt = ( 16/27 ) [pi] [ 3 h^{2}dh/dt ].^{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. - 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] r
^{2}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 m^{3}/min
from a shallow concrete conical reservoir (vertex down) of base radius 45 m
and height 6 m.

- How fast (cm/min) is the water level falling when the water is 5 m deep?
- 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] r^{2} h and we are told that water is flowing
out of the reservoir at a rate of 50 m^{3}/min, which means that
dV/dt = -50 (because the volume is *decreasing*).

- 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] r^{2}h and r = ( 15/2 ) h, we haveV = ( 1/3 ) [pi] [ ( 15/2 ) h ] Evaluating this when h = 5 given that dV/dt = -50, we have -50 = ( 225/12 ) [pi] [ 3 ( 5 )^{2}h = ( 225/12 ) [pi] h^{3}

dV/dt = ( 225/12 ) [pi] [ 3 h^{2}dh/dt ]^{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. - 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.

- How fast is the boat approaching the dock when 10 ft of rope are out?
- 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.

- By the Pythagorean Theorem, we have R
^{2}= H^{2}+ 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. HencedH/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*). - Let A denote the angle between the rope and the dock, so tan A = H / 6.
Thus [ sec
^{2}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, sodA/dt = [ cos which says that the angle is changing at the rate of -0.15 radians/sec when R = 10.^{2}A ] ( 1/6 ) [ dH/dt ] = [ 6/10 ]^{2}( 1/6 ) [ -2.5 ] = -0.15

**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.

- How fast is the top of the ladder sliding down the wall then?
- At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
- 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
G^{2} + W^{2} = 13^{2} and we are told that when
G = 12, dG/dt = 5.

- Differentiating G
^{2}+ W^{2}= 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 finddW/dt = ( -G / W ) dG/dt = [ ( -12 ) / ( 5 ) ] [ 5 ] = -12, i.e., the ladder is sliding*down the wall*12 ft/sec. - 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 ft^{2}/sec. - Let a be the angle between the ladder and the ground. Then tan a =
W / 13 = ( 1/13 ) W, so [ sec
^{2}a ] da/dt = ( 1/13 ) dW/dt. When G = 12 and W = 5, sec a = hyp/adj = 13/12, soda/dt = [ cos a ] and the angle is^{2}( 1/13 ) dW/dt = [ 12/13 ]^{2}( 1/13 ) [ -12 ] = -0.7865*decreasing*at the rate of 0.7865 radians/sec.

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

- 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?
- 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
y^{2} = 90^{2} + ( 90 - x )^{2}. In both parts of
the problem, we want to find how the distance y changes, i.e., what is dy/dt.

- When x = 30, we are told that dx/dt = 16. Differentiating y
^{2}= 90^{2}+ ( 90 - x )^{2}with respect to time, we find 2 y dy/dt = 2 ( 90 - x ) [ -dx/dt ], sody/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. - 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
x^{2} + xy + y^{2} = 7.

- Find the points where the tangent is parallel to the x-axis.
- Find the points where the tangent is parallel to the y-axis.
- 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 ).

- 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 x
^{2}+ xy + y^{2}= 7, which we solve by substitution: x^{2}+ x ( -2x ) + ( -2x )^{2}= x^{2}- 2x^{2}+ 4x^{2}= 3x^{2}, so x^{2}= 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 x^{2}+ xy + y^{2}= 7 where the tangent line is paralle to the x-axis are (1.5275,-3.055) and (-1.5275,3.055). - 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 x
^{2}+ xy + y^{2}= 7 and x = -2y, which we do by substitution: ( -2y )^{2}+ ( -2y ) y + y^{2}= 3y^{2}= 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). - For this last part, we need to find the points where the curve crosses
the x-axis, which happens when y = 0 so that x
^{2}= 7, i.e., x =__+__2.64575. At these points, the tangents to the curve have slope:m which are the same so the tangent lines are parallel and both of slope m = -2._{(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

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

- y = x
^{ln x} - y = x
^{( 1 / ln x )} - y = [ x ( x
^{2}+ 1 )^{1/2}] / [ ( x + 1 )^{2/3}]

**Solution**:

- Setting y = x
^{ln x}, take the natural log of both sides to get ln y = ln [ x^{ln x}] = ( ln x ) ln [ x ] = ( ln x )^{2}. Thus ( 1 / y ) dy/dx = 2 ( ln x ) [ 1 / x ], sody/dx = y [ 2 ( ln x ) / x ] = [ x ^{ln x}] [ ( 2 ln x ) / x ] = 2 ( ln x ) x^{( ln x ) - 1}. - 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. - Taking ln, we have ln y = ln x + ( 1/2 ) ln( x
^{2}+ 1 ) - ( 2/3 ) ln( x + 1 ), so ( 1 / y ) dy/dx = 1/x + ( 1/2 ) [ 1 / ( x^{2}+ 1 ) ] [ 2x ] - ( 2/3 ) [ 1 / ( x + 1 ) ] [ 1 ] = 1/x + [ x / ( x^{2}+ 1 ) ] - [ 2 / 3( x + 1 ) ]. Hencedy/dx = y [ 1/x + ( x / ( x ^{2}+ 1 ) ) - ( 2 / 3( x + 1 ) ) ] = [ x ( x^{2}+ 1 )^{1/2}] / [ ( x + 1 )^{2/3}] [ 1/x + ( x / ( x^{2}+ 1 ) ) - ( 2 / 3( x + 1 ) ) ].

**Problem 10**: Consider the following four functions:
f(x) = e^{x} - 1,
g(x) = x^{3} + x,
h(x) = ln( x + 1 ),
j(x) = sin x.

- Find the linearizations of f, g, h and j near x = 0.
- Given your answer to part (a), what is special about those four particular functions that made the result unusual?
- 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.

- The linearization of f(x) = e
^{x}- 1 is thereforeL because f'(x) = e_{f}(x) = f(0) + f'(0) x = [ 0 ] + [ e^{0}] x = x^{x}. The linearization of g(x) = x^{3}+ x isL because g'(x) = 3x_{g}(x) = g(0) + g'(0) x = [ 0 ] + [ 3 ( 0 )^{2}+ 1 ] x = x^{2}+ 1. The linearization of h(x) = ln( x + 1 ) isL because h'(x) = 1 / ( x + 1 ). The linearization of j(x) = sin x is_{h}(x) = h(0) + h'(0) x = [ 0 ] + [ 1 / ( ( 0 ) + 1 ) ] x = xL since j'(x) = cos x._{j}(x) = j(0) + j'(0) x = [ 0 ] + [ cos 0 ] x = x - 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. - The linear approximation L
_{f}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 L_{g}is within the allowed error of__+__0.5 for x roughly between -0.8 and 0.8. The linearization L_{h}is within the allowed error of__+__0.5 for x roughly between -0.7 and 1.4. Finally, the linearization L_{j}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) = x^{3}+ 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

**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] r^{3} to find dV = ( 4/3 ) [pi] [ 3 r^{2}
dr ]. Thus the approximate volume of paint required is the value of dV
when r = 100 and dr = 0.02, i.e.,

**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] r^{2}, 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] r^{3} within
1%?

**Solution**: We want dS = 8 [pi] r dr to be less than or equal
to S/100 = [ 4 [pi] r^{2} ] / 100. Thus we want 8 [pi] r dr __<__
[ 4 [pi] r^{2} ] / 100, so dr __<__ [ 4 [pi] r^{2} ] /
[ 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,

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 r^{4}, we have dV = k [ 4
r^{3} dr ] = 4k r^{3} dr. Then dV / V = [ 4k r^{3}
dr ] / [ k r^{4} ] = [ 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).

- Find equations for the tangent and normal to the cissoid of Diocles
y
^{2}( 2 - x ) = x^{3}at (1,1). - The line that is normal to the curve x
^{2}+ 2xy - 3y^{2}= 0 at (1,1) intersects the curve at what other point? - Show that if it is possible to draw three normals from the point (a,0)
to the parabola x = y
^{2}, 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**:

- To begin with, we want to find the value of dy/dx at the point (1,1) on
the cissoid of Diocles, y
^{2}( 2 - x ) = x^{3}. Implicitly differentiating, we obtain y^{2}[ -1 ] + ( 2 - x ) [ 2y dy/dx ] = 3x^{2}. Solving for dy/dx, we have 2y ( 2 - x ) dy/dx = 3x^{2}+ y^{2}, so dy/dx = [ 3x^{2}+ y^{2}] / [ 2y ( 2 - x ) ]. Thus, at (1,1), the tangent line to the cissoid has slopedy/dx = [ 3 (1) Hence the equation of the^{2}+ (1)^{2}] / [ 2 ( 1 ) ( 2 - ( 1 ) ) ] = [ 3 + 1 ] / [ 2 ] = 2.*tangent line*isy - 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 isy - 1 = ( -1/2 ) ( x - 1 ). - Find the equation of the normal to x
^{2}+ 2xy - 3y^{2}= 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 isy - 1 = -1 ( x - 1 ). To find the other point on the curve x^{2}+ 2xy - 3y^{2}= 0 where the normal y - 1 = -x + 1 (or y = -x + 2) intersects the curve, substitute y = 2 - x into the equationx and solve for x:^{2}+ 2xy - 3y^{2}= x^{2}+ 2x( 2 - x ) - 3 ( 2 - x )^{2}= 0x 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 x^{2}+ 4x - 2x^{2}- 3 ( 4 - 4x + x^{2}) = -12 + 16x - 4x^{2}= -4 ( x^{2}- 4x + 3 ) = -4 ( x - 1 ) ( x - 3 ) = 0^{2}+ 2xy - 3y^{2}= 0. - Let's first find what the slope of the tangent line to the curve x =
y
^{2}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 (y^{2},y) on x = y^{2}, the tangent line has slope 1 / [ 2y ]. At that same point, the line joining the point (a,0) to (y^{2},y) has slopem = [ y - 0 ] / [ y For these lines to be perpendicular, and thus for the segment joining (a,0) to (y^{2}- a ] = y / [ y^{2}- a ].^{2},y) to be normal to x = y^{2}, the product of the slope m = y / [ y^{2}- a ] with the tangent line slope 1 / [ 2y ] must be equal to -1:-1 = [ 1 / ( 2y ) ] [ y / ( y so y^{2}- a ) ] = 1 / [ 2 ( y^{2}- a ) ],^{2}- a = -1/2. Thusa = 1/2 + y for all y. Thus, for there to be^{2}__>__1/2*three*points where the normals to x = y^{2}pass through (a,0), for two of the points we must have y different from 0 and thus we must have a = 1/2 + y^{2}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 botha = 1/2 + y (the first requirement is to ensure that the segments joining (a,0) to (y^{2}*and*[ y / ( y^{2}- a ) ] [ -y / ( y^{2}- a ) ] = -1^{2},__+__y) are normal to x = y^{2}and the second to ensure that these segments are perpendicular to each other). The second equation becomes y^{2}= ( y^{2}- a )^{2}= y^{4}- 2ay^{2}+ a^{2}upon simplifying and clearing denominators. If we now use the first, in the form y^{2}= a - 1/2, to replace y^{2}and y^{4}we gety Thus a - 1/2 = 1/4, so a = 3/4 is our solution.^{2}= y^{4}- 2ay^{2}+ a^{2}

a - 1/2 = ( a - 1/2 )^{2}- 2a ( a - 1/2 ) + a^{2}= ( a^{2}- a + 1/4 ) - 2a^{2}+ a + a^{2}= 1/4.

**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
s^{3}, 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 = s^{3} and its surface
area is S = 6s^{2}, since it has six sides each of area s^{2}.
Thus, we are given that dV/dt = -k S = -k ( 6s^{2} ), but we can
also calculate dV/dt = [ dV/ds ] [ ds/dt ] = [ 3 s^{2} ] ds/dt by
the chain rule from the formula V = s^{3}. Thus 3s^{2} ds/dt
= -k ( 6s^{2} ), so ds/dt = -2k. That is, the side length is
decreasing at a constant rate of 2k units per hour.
Thus, if s_{0} is the the initial side length of the ice cube, the
length of its side one hour later is s_{1} = s_{0} - 2k.
Therefore

**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

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,

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

**Problem 1**: Suppose that the distance an aircraft travels along
a runway before takeoff is given by D = ( 10 / 9 ) t^{2}, 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 ) t^{2}.
Thus v(t) = dD/dt = d/dt [ ( 10/9 ) t^{2} ] = ( 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) = x^{3} -
6x^{2} + 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 [ x^{3} - 6x^{2} + 15x ]
= 3x^{2} - 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 = 10^{6} + 10^{4} t -
10^{3} t^{2}. 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 [ 10^{6} + 10^{4} t - 10^{3} t^{2} ]
= 10^{4} - 10^{3} [ 2t ]. Thus, the growth rates are

- t = 0: db/dt = 10
^{4}- 10^{3}[ 0 ] = 10^{4} - t = 5: db/dt = 10
^{4}- 10^{3}[ 2(5) ] = 0 - t = 10: db/dt = 10
^{4}- 10^{3}[ 2(10) ] = -10^{4}

**Problem 4**: Define f(1) in a way that extends f(s) =
[ s^{3} - 1 ] / [ s^{2} - 1 ] to be continuous at s = 1.

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

- f(1) is defined
- the limit as s approaches 1 of f(s) exists
- the value of f(1) is equal to the value of the limit as s approaches 1.

**Problem 5**: For what value of A is the function f(x) defined
by f(x) = x^{2} - 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
x^{2} - 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 x^{2} -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 x^{2} - 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 x^{2} - 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 x^{2} - 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 x^{2} - 1 and (8/3)x at x = 3) to be equal. Yet
the left-hand derivative is the value of d/dx[ x^{2} - 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.