Calculus with Analytic Geometry I
Exam 10–Take Home part
Textbook, Section 4.7, Exercises #22, 30, 32, 38, 48, 56, 70, 76
1. (# 22) Find, correct to two decimal places, the coordinates of the point on the curve y = sin x that is closest
to the point (4, 2).
Solution. We find the point (x, y) on the curve that minimizes the square of the distance D = (x − 4)2 +
(y − 2)2 . The constraint on (x, y) is that y = sin x, so as a function of x,
D(x) = (x − 4)2 + (sin x − 2)2 .
The domain in which we have to consider this function is all of the real line; x could be anywhere.
We begin finding the critical points of D; we compute d0 (x) and set it to 0.
D0 (x) = 2(x − 4) + 2(sin x − 2) cos x = 0.
This equation can be rewritten as (2 − sin x) cos x = x − 4. This is not an easy equation to solve, which is why
the book has the graphic calculator or computer symbol attached to it. Those of you who get stymied, will
you bother to ask for help? But thanks to the magic of Wolfram alpha, it gets to be quite easy to complete
the exercise. First I will draw a graph to get an idea of how many solutions there might be. Notice that the
left hand side is always between −2 and 3 (because 2 − sin x is always ≥ 1 and ≤ 3, so with −1 ≤ cos x ≤ 1,
the worst case scenario is that cosx = −1 and 2−sin x = 2, so the product is −2, or cos x = 1 and 2−sin x = 3
so the product is 3), so that we must have −2 ≤ x − 4 ≤ 3, or 2 ≤ x ≤ 7. So if there is a solution, it will be
in the interval [2, 7]. This can help us with the graph. What I will have Wolfram alpha do is plot the two
curves y = (2 − sin x) cos x and y = x − 4 in the interval [2, 7] and see how many times they intersect; if any
at all. The result is
We see clearly∗ that there are three solutions. So next we ask our friend W. alpha to find them, by inputting
solve (2 − sin x) cos x = x − 4. From the graph we already have a good idea of what the solutions should be,
close to 2.65, 5 and 6. W. alpha’s answer is
x ≈ 2.65119, 5.09625, 6.12841
We need to decide which is a minimum, if any. Let’s be logical about it. Use our head. Think of x as being
far away to the left, very negative,and now it begins to move to the right. The distance between (x, sin x)
and (4, 2) has to be decreasing. This means that 2.65119 is either a local minimum or an inflection. Let us
check this using the second derivative test, keeping in mind that the x-value is only approximate. We have
D00 (x) = 2(1 + cos2 x − sin2 x + 2 sin x); D00 (2.65119) = 4.9966 . . .
D00 cam out sufficiently positive that we can rule out an inflection point; we must have a local minimum.
After this, D increases; this makes 5.09625 either a local max, or an inflection. Computing,
D00 (5.09625) = −3.14789 . . . ,
∗ Well, one has to be careful. The curve could have a number of little oscillations that are too small to show up on the graph. If this
were a life and death project, one might want to run a few more tests.
2
so we have a local maximum. This forces the third critical point to be a local minimum, since D must increase
as x → ∞. It is now quite evident that whichever one of D(2.65119), D(6.12841) is smaller, is the minimum
value. We have
D(2.65119) = 4.157 . . . ,
D(6.12841) = 0.17 . . .
The conclusion is that the first critical point is the one we are looking for.
We could also have graphed D itself; here is how the graph looks.
This confirms our calculations; the critical point at approximately 2.65 is where the absolute minimum occurs.
Now sin(2.65119) = 0.47098. Rounding off to the required two decimal places, the answer is (2.65, 0.47).
2. (#30) A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible
volume of such a cylinder.
Solution. We must assume that “inscribed” means the the base of the cylinder is contained in the base
of the cone and geometrical intuition shows that to get the largest cylinder we should have the center of the
bases of cylinder and cone be the same. Let x be the radius of the cylinder, y its height. If we cut by a
vertical plane through the common axis, we get the following picture.
3
The volume of the cylinder is V = πx2 y. We need to relate x and y, which is easily done by similar
triangles. In the profile pictured, the triangle above the green rectangle is similar to the big triangle. We get
(h − y)/h = x/r. Solving, y = h − (hx/r). Then
hx
x3
V (x) = πx2 h −
= πh x2 −
.
r
r
3x2
The domain of this function is 0 ≤ x ≤ r. Now V 0 (x) = πh2 2x −
; setting to 0 we have two solutions,
r
x = 0 (an end-point) and x = 2r/3. Because of the Extreme Value Theorem, there is a maximum. Because
interior maxima are achieved at critical points, the maximum value of the volume occurs for x = 0, 2r/3, or
x = h. Now V (0) = 0, V (2r/3) = 4πhr2 /27, V (h) = 0. The maximum value of the volume is V =
4πhr2
.
27
3. (#32) A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the
semicircle is equal to the width of the rectangle). If the perimeter of the window is 30 ft., find the dimensions
of the window so that the greatest possible amount of light is admitted.
Solution.
Drawing a picture is indicated.
1 x 2
πx2
With the variables from the picture, the area of the window is A = xy + π
= xy +
. The constraint
2
2
8
is 2y + x + πx/2 = 30. Solving for y, substituting into the area, simplifying a bit,
A(x) = 15x −
π+4 2
x .
8
π+2
The domain of this function is a bit messy; clearly x ≥ 0. We also must have y = 15 −
x ≥ 0, which
4
60
60
π
+
4
forces x ≤
. So 0 ≤ x ≤
. Differentiating A we get A0 (x) = 15 −
x. Setting A0 (x) = 0, there
π+2
π+2
4
is a single cortical point, namely x = 60/(π + 4). To justify it is the maximum, for once we can compute A00 ;
A00 (x) = −(π + 4)/4 < 0. Because 60/(π + 4) is the ONLY critical point in the interval, and it is a relative
max, it is where the max occurs. The dimensions are
x=
60
,
π+4
y=
30
.
π+4
4
4. (# 38) A fence 8 ft. tall runs parallel to a tall building a ta distance of 4 ft from the building. What is the
length of the shortest ladder that will reach from the ground over the fence to the wall of the building?
Solution.
Here’s a picture.
The ladder is depicted in red. The square of its length is S = x2 + y 2 ; I’ll minimize the square of the length
rather than the length. From similar triangles, y/x = 8/(x − 4); thus y = 8x/(x − 4) and
S = x2 +
64x2
.
(x − 4)2
The domain of this function is 0 ≤ x < ∞. Differentiating and setting to 0 we get
S 0 (x) = 2x −
512x
= 0;
(x − 4)3
solving, x = 0 and x = 4(1 + 22/3 ). The only relevant solution is the second one; x = 0 is not in the domain.
This time I’ll use the first derivative test to justify that we found a minimum. Notice that limx→4+ S 0 (x) =
−∞, limx→∞ S 0 (x) = ∞. For this to happen, S 0 must be negative close and the the right of 4, positive
for large values of x. This means, S 0 goes from negative to positive at the critical point, thus it is a local
minimum. Being the ONLY critical point, there ia an absolute minimum at that point. Now, at this value of
x,
3
64
64
2/3 2
2/3
=
16(1
+
2
)
1
+
=
16
1
+
2
.
S(x) = x2 1 +
(x − 4)2
32 · 21/3
The length of the ladder is the square of this quantity, thus the length of the shortest ladder is
3/2
L = 4 1 + 22/3
≈ 16.65 ft.
5. (#48) A woman at a point A on the shore of a circular lake with radius 2 mi wants to arrive at the point C
diametrically opposite A on the other side of the lake in the shortest possible time (see figure). She can walk
at the rate of 4 mi/h and row a boat at 2 mi/h. How should she proceed?
Solution.
Lets add some data to the picture:
5
Using a bit of high school geometry (or, what should be high school geometry) the central angle subtending
the arc BC is twice the angle θ. By the law of cosines, and high school trigonometry
|BC|2 = r2 + r2 − 2r · r cos(π − 2θ) = 2r2 (1 − cos(π − 2θ)) = 2r2 (1 + cos 2θ) = 4r2 cos2 θ,
where r = 2 is the radius of the circular lake. That is |BC|2 = 16 cos2 θ, thus |BC| = 4 cos θ. The length of
the arc BC is 2θr = 4θ. The total time the woman would take if she rows to B and then walks, is
T (θ) =
4 cos θ 4θ
+
= 2 cos θ + θ.
2
θ
The domain of T is the interval [0, π/2]. Differentiating and setting to 0 gives T 0 (θ) = −2 sin θ + 1 = 0, thus
sin θ = 1/2, hence θ = π/6. The minimum (and the maximum) value of T has to occur at one of 0, π/6, π/2.
We see that
√
π
π
π
T (0) = 2, T (π/6) = 3 + ≈ 2.26, T ( ) = ≈ 1.57.
6
2
2
The minimum occurs for θ = π/2, which means that the woman should walk all the way.
6. (#56) What is the smallest possible area of the triangle that is cut off by the first quadrant and whose
hypotenuse is tangent to the parabola y = 4 − x2 at some point?
Solution.
The picture shows one such triangle.
The variables to be used are the coordinates of the point of tangency, namely (a, b). Since y 0 = −2x, the slope
of the tangent line at (a, b) is −2a. The equation of the tangent line is y − b = −2a(x − a). But b = 4 − a2 ,
so it is
y − 4 + a2 = −2a(x − a), simplifying y = −2ax + 4 + a2 .
The y and x intercepts of this line are, respectively, 4 + a2 and (4 + a2 )/2a. The x-intercept is the base, the
y-intercept the height, of the triangle, the area is thus
A(a) =
(4 + a2 )2
4
a3
= + 2a + .
4a
a
4
6
The domain is 0 < a ≤ 2 (b = 4 − a2 ≥ 0 forces a2 ≤ 4, so a ≤ 2). Now setting A0 (a) = 0:
0 = A0 (a) = −
3
4
3a4 + 8a2 − 16
+ 2 + a2 =
.
2
a
4
4a2
Thus 3a2 + 8a2 − 16 = 0, a quadratic equation in a2 ; with the solutions
a2 = (−8 ± 16)/6. a2 cannot be
√
2
negative, so the only valid solution is a = 8/6 = 4/3, so a = 2/ 3, which is in the domain. Now A0 (a) is
seen to be negative near 0, positive for large values of a, thus we have a minimum (single critical point in the
√
32 3
.
interval criterion!). The corresponding area is A =
9
7. (#70) A steel pipe is carried down a hallway 9 ft wide. At the end of the hall there is a right angled turn into
a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around
the corner?
Hint: Suppose L is the length of the slanted line in the picture, the one forming an angle of θ with the first
hallway. A bit of trigonometry might allow you to relate this length to θ. The longest the pipe could be
might perhaps be the largest possible value of L? You might also notice that 0 ≤ θ ≤ π/2.
Solution. The so called hint has a little mistake in it. The longest the pipe can be is the shortest of all
the lengths L of the type in the picture, because if it is to go through the bend, it can’t be longer than this
shortest length.
Let us call x the length of the part of L in the wider corridor, y the part in the narrower corridor, so that
L = x + y. Now x = 9/ sin θ, y = 6/ cos θ (see picture).
Thus L(θ) =
9
6
+
= 9 csc θ + 6 sec θ the domain being 0 < θ < π/2. Now
sin θ cos θ
9 cos θ 6 sin θ
6 sin3 θ − 9 cos3 θ
+
=
.
cos2 θ
sin2 θ
cos2 θ sin2 θ
p
√
Setting L0 (θ) = 0, we get 6 sin3 θ = 9 cos3 θ, thus tan θ = 3 9/6 = 3 1.5. Is the value at this θ a minimum?
The best way of seeing that it is, is using an intuitively clear fact, which is also an easy consequence of
the Extreme Value Theorem, and the fact that interior extrema are achieved at critical points. Here is the
intuitively clear fact:
L0 (θ) = −9 csc θ cot θ + 6 sec θ tan θ = −
Let f be defined and continuous on an interval (a, b) (that is f (x) is defined for all x such that a < x < b,
and f is continuous at all points of (a, b)). If limx→a+ f (x) = ∞ and limx→a− f (x) = ∞, then f assumes a
7
minimum value in (a, b); that is, there is a point x0 in (a, b) such that f (x0 ) is the absolute minimum value
of f . If f is differentiable at that point, then f 0 (x0 ) = 0.
The previous result is true also if a = −∞, in which case limx→a+ f (x) is replaced by limx→−∞ f (x) and/or
if b = ∞; then limx→b+ f (x) becomes limx→∞ f (x).
Noticing that in our case limθ→0+ L(θ) = limθ→(π/2)− L(θ) = ∞, the principle just mentioned guarantees a
point in (0, π/2) where an absolute minimum occurs; it also guarantees that at that point L0 is zero. Well,
there is only one point where L0 is zero; L at that point must be the absolute minimum. Since we are√asked
for the length of the longest pipe (which coincides √
with the smallest value of L) we have to calculate L( 3 1.5).
This means computing the cosine and sine of this 3 1.5 and I will avoid using approximations and calculators
until it becomes absolutely necessary. So we set up the standard triangle.
The tangent of the pictured angle θ is
√
3
1.5. We see that
cos θ = √
1
1 + 1.52/3
,
sin θ = √
1.51/3
1 + 1.52/3
,
the shortest L, and the longest pipe, will have length
√
p
p
9 1 + 1.52/3
1.5
2/3
2/3
L =
= 6 1 + 1.5
+ 6 1 + 1.5
+1
1.51/3
1.51/3
3/2
p
= 6 1 + 1.52/3 1.52/3 + 1 = 6 1.52/3 + 1
.
And the answer is
3/2
6 1.52/3 + 1
≈ 21.07 ft.
8. (#76) The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that
convey blood from the heart to the organs and back to the heart. This system should work so as to minimize
the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the
resistance of the blood is lowered. One of Poiseuille’s laws gives us the resistance R of the blood as
R=C
L
r4
where L is the length of the blood vessel, r is the radius, and C is a positive constant determined by the
viscosity of the blood (Poiseuille established this law experimentally, but it also follows from equation 2 in
section 4 of Chapter 8). The figure shows a main blood vessel with radius r1 branching at an angle θ into a
smaller vessel with radius r2 .
8
(a) Use Poiseuille’s law to show that the total resistance of the blood along the path ABC is
a − b cot θ b csc θ
R=C
+
r14
r24
where a and b are the distances shown in the figure.
Solution. We have to assume that resistance is additive. Consider the triangle with vertices at B, C,
and the third vertex at the right end of the main blood vessel. The length from point B to that third
vertex is b/ tan θ = b cot θ. The distance from A to B is thus a − b cot θ, making the resistance for that
portion of the main vessel equal to C(a − b cot θ)/r14 . The length from B to C is b/ sin θ = b csc θ, so the
resistance in that secondary vessel will be R2 = Cb csc θ/r24 . Adding up R1 + R2 , we get precisely the R
given above.
(b) Prove this resistance is minimized when
cos θ =
r24
.
r14
Solution.
The assumption here is that a, b are constants; otherwise we’d have no idea how to proceed. This means
that the smallest θ can be is when its tangent is b/a; the largest is π/2. That is, the domain of R is
arctan(b/a) ≤ θ ≤ π/2.
We find the critical points of R. Differentiating (the C in front is a nuisance, by the way)
1
1
Cb
1
cos θ
2
R0 (θ) = C
b
csc
θ
−
b
csc
θ
cot
θ
=
−
.
r14
r24
r24
sin2 θ r14
Setting to cos θ = r24 /r14 follows. Is it a minimum? I took a point away for not justifying that
a minimum or maximum was achieved. In the final, I will probably penalize you more
for lack of justification! The answer to the question, is it a minimum is actually “not necessarily.”
Looking at the expression for R0 (θ) it is clear that R0 (θ) > 0 if θ is to the right of the point; for example
when θ = π/2, R0 (π/2) = Cb/r14 > 0. It is also easy to see that R0 (θ) < 0 to the left of our presumed
minimum point. The big problem is: Is θ such that cos θ = r24 /r14 in the domain of R?. Since cos
decreases in [0, π/2] we must have r24 /r14 ≤ than the value of cos on the left end point of the domain;
that is,
r24
b
a
≤ cos(arctan ) = √
.
4
2
r1
a
a + b2
Nothing is said about the relation between a, b and the radii of√the vessels. Presumably for the same
radii r2 , r1 we might √
have a very large value of b, making a/( a2 + b2 very small, so small that the
4 4
inequality r2 /r1 ≤ a/ 12 + b2 does not hold. In that case the minimum is achieved for θ = arctan(b/a).
Our beloved author does not seem to have considered this possibility (or explained why it can’t happen).
9
(c) Find the optimal branching angle (correct to the nearest degree) when the radius of the smaller blood
vessel is two-thirds the radius of the larger vessel.
√
Solution. Assuming that r24 /r14 ≤ a/ a2 + b2 , the minimum occurs for cos θ = r24 /r14 , which in this
case equals (2r1 /3)4 /r14 = 16/81. In degrees the angle is
θ = arccos
16
≈ 78.61◦ .
81
(In the text there is now a picture of actual blood vessels.)