Calculus with Analytic Geometry IExam 10, Take Home –Friday, November 8, 2013

Solutions.

All exercises are from Section 4.7 of the textbook.

1.

Solution. The picture suggests using the angle θ as variable; θ = 0 means she rows all the way, θ = π/2,she walks all the way. There are several ways of getting all the distances; here is one. Let O be the center ofthe lake and consider the isosceles triangle AOB. The base angles must be equal, thus the angle AOB (angleat O) is π − 2θ. The length |AB| is then given by

|AB|2 = 22 + 22 − 2 · 2 · 2 cos(π − 2θ) = 8 + 8cos(2θ) = 8(1 + cos2 θ − sin2 θ) = 16 cos2 θ,

so that |AB| = 4 cos θ. Concerning the length of the arc from B to C, its length equals the radius times thesubtended angle COB. This angle is again easily seen to be 2θ; thus the length of the arc is 4θ. We now seethat the time to minimize is

T (θ) =4 cos θ

2+

4θ

4= 2 cos θ + θ, 0 ≤ θ ≤ π

2.

We find critical points, if any. We have T ′(θ) = −2 sin θ+ 1; setting it to 0 we get sin θ = 1/2. In the interval[0, π/2], sin θ = 1/2 only for θ = π/6. The easiest way of checking is to just evaluate T at the critical pointand at the endpoints of the interval. We get

T (0) = 2, T (π/6) = 2 cos(π/6) + π/6 =√

3 +π

6≈ 2.26, T (π/2) =

π

2≈ 1.57

To minimize time, she should walk all the way.

2.

Here is a picture to go with the problem.

A convenient variable is x, where (x, 4 − x2) is the point of tangency. But we have to be careful in how wewrite the equation of the line, so for just a moment it may be better to denote the point of tangency by

(x0, 4 − x20). The value of the derivative at this point isd(4− x2)

dx

∣∣∣∣x0

= −2x0. The equation of the tangent

line is theny − (4− x20) = −2x0(x− x0), or y = −2x0x+ x20 + 4.

From this we get at once the values of the x and y intercepts of the tangent line; they are, respectively, x0

2 + 2x0

and x20 + 4. These are also the lengths of the legs of the triangle so that, writing again x for x0, the area tominimize is

A(x) =1

2

(x

2+

2

x

)(4 + x2) =

1

4x3 + 2x+

4

x.

The domain in which we have to consider this function is (0,∞). We see that A′(x) =3

4x2+2− 4

x2; setting to

0 and multiplying by 4x2 (which changes nothing since x > 0) we get 3x4 + 8x2− 16 = 0. This is a quadraticequation in x2 with roots −4 and 4/3. Discarding the impossible root −4, we are left with x2 = 4/3, thusx = ±2/

√3. The only critical point in the domain of relevance is x = 2/

√3. Now A′′(x) = 3

2x+ 8x3 > 0 when

x > 0, so that we have a local minimum, hence an absolute minimum since it is the only critical point. Theminimum area is thus

A

(2√3

)=

32√

3

9.

3.

Solution. I’ll denote by |AB| the length of the segment AB. The length of the rope is L = |PQ| csc θ1 +|ST | csc θ2. We have to relate θ2 with θ1. We can use that

tan θ2 =|ST ||RT |

so that a standard trigonometry exercise shows that csc θ2 =

√|RT |2 + |ST |2|ST |

. Now

|RT | = |QT | − |QR| = |QT | − |PQ| cot θ1,

so writing now θ for θ1 we get for the length of the rope

L = |PQ| sec θ +√

(|QT | − |PQ| cot θ)2 + |ST |2

But this gives a really nasty derivative to minimize, so let us try to find instead the point R that minimizesthe length of the rope. I’ll introduce for this purpose the variable x = |QR| and now we can express thelength L of the rope in the somewhat simpler form

L(x) =√|PQ|2 + x2 +

√|ST |2 + (|QT | − x)2.

The domain of the variable x is 0 ≤ x ≤ |QT |. Then

L′(x) =x√

|PQ|2 + x2− |QT | − x√

|ST |2 + (|QT | − x)2

Setting to 0, and solving for x on gets x =|QT | |PQ||PQ|+ |ST |

. Is this the x that minimizes L?. We notice that

L′(0) = − |QT |√|ST |2 + |QT |2

< 0, L′(|QT |) =|QT |√

|ST |2 + |QT |2> 0;

since L′ is only zero at x, that means L′ is negative to the left of the critical point, positive to the right; thecritical point is a local minimum and, being the only critical point in the interval, the point at which theabsolute minimum occurs.

With this value of x = |QR| we get

tan θ1 =|PQ|x

=|PQ|+ |ST ||QT |

,

tan θ2 =|ST ||QT | − x

=|PQ|+ |ST ||QT |

.

The angles are equal.

NOTE: There is a clever way of solving this problem without appealing to calculus. Can you find it?

4.

Solution. The amount of water will be maximum when the cross-sectional area is maximum. The crosssection is a trapezoid of bases of lengths (in cm) 10 and 10 + 2 · 10 cos θ, height 10 sin θ. The area is thus

A(θ) = 100(1 + cos θ) sin θ.

the domain∗ of A is 0 ≤ θ ≤ π/2. Now

A′(θ) = 100(− sin2 θ + (1 + cos θ) cos θ) = 100(cos θ + cos2 θ − sin2 θ).

To be able to solve for θ after setting A′(θ)to0, it is a good idea to replace sin2 θ by 1 − cos2 θ. With thissubstitution, we get

A′(θ) = 100(2 cos2 θ + cos θ − 1).

Setting A′(θ) to 0 yields a quadratic equation in cos θ that can be solved by the quadratic formula:

2 cos2 θ + cos θ − 1 = 0, so cos θ =−1± 3

4,

so θ = 1/2 or θ = −1. The second solution can’t occur in the given domain, so cos θ = 1/2, hence θ = π/3 isthe only critical point in the domain. Now

A(0) = 0, A(π/3) = 75√

3 ≈ 129.9, A(π/2) = 100.

The maximum value occurs when θ =π

3.

5.

∗One can stop the domain at π/2 since it is obvious that a larger angle will give a worse cross-sectional area. But one can alsoconsider a somewhat larger domain. However, one cannot take θ all the way to π because once θ = 5π/6 (150◦) the two sides of thegutter touch, the cross section is an equilateral triangle, and it doesn’t make sense to keep bending the sides.

Solution. This exercise is way easier than it seems. It is a bit a lesson in anatomy.

(a) A bit of trigonometry shows that the distance from B to the end of the horizontal blood vessel is b cot θ.The distance from A to B is thus a− b cot θ. Trigonometry also shows that the distance from B to C isb csc θ. So assuming (as one has to, to get the required answer) that resistance is additive, the resistancefrom A to B is in fact

R = C

(a− b cot θ

r41+b csc θ

r42

).

(b) I assume that the full question is: Given a major blood vessel (an artery, perhaps), a sub-vessel hasto branch off to get blood to a point C at distance b from the major vessel, and it should branch offat a point at a distance less than a from the point nearest to C. So we do the usual. First of all,let us determine the domain of R as a function of θ. It should be clear that θ = 0 cannot work. Infact, the smallest value of θ has to have a tangent of b/a and the largest is π/2. The domain of R isarctan(b/a) ≤ θ ≤ π/2. Differentiating,

R′(θ) = C

(+b csc2 θ

r41− b csc θ cot θ

r42

)= Cb csc θ

(csc θ

r41− cot θ

r42

);

setting to 0, since csc θ is never 0, we getcsc θ

r41− cot θ

r42= 0, thus

cot θ

csc θ=r42r41

and, since cot θ/ csc θ = cos θ,

we see that R′(θ) = 0 if and only if cos θ = r42/r41. Is this in the domain? Since we can assume that a is

fairly large compared to b, so arctan b/a is close to 0, the answer is probably yes. Justifying that this isa minimum (and not a maximum) is perhaps harder than usual, but not impossible. Here is one way.While the domain of R might not go all the way to θ = 0, the function R(θ) is define in (0, π) . If we

replace cot, csc by their expressions in terms of sin, cos, the derivative becomes

R′(θ) = Cb1

sin2 θ

(1

r41− cos θ

r42

),

showing that the sign of R′(θ) is the same as the sign of1

r41− cos θ

r42. Now

1

r41− cos θ

r42

∣∣∣∣θ=0

=1

r41− 1

r42< 0

because r2 < r1, while1

r41− cos θ

r42

∣∣∣∣θ=π

2

=1

r41> 0.

The derivative goes from negative to positive, thus we have a minimum.

(c) We have to find θ such that

cos θ =

(23r1)4

r41=

16

91.

Thus θ = arccos 1691 ≈ 80◦.