Solutions to Final Examination in Calculus I, Study notes of Calculus

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Solutions to Final Examination

V63.

Calculus I

December 19, 2008

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1. (10 Points) Compute the following derivatives. Please leave your answers unsimplified.

(i)

d dx

(ln (x sin x + 1 ))

Solution. We need the chain rule and the product rule:

d dx

(ln (x sin x + 1 )) =

x sin x + 1

d dx

(x sin x + 1 ) = x cos(x) + sin(x) x sin(x) + 1

N

(ii)

d dx

arctan

x + 1 ex^ + 1

Solution. Again by the chain rule,

d dx

arctan

x + 1 ex^ + 1

x+ 1 ex^ + 1

) 2 ·^

d dx

x + 1 ex^ + 1

x+ 1 ex^ + 1

( 1 + ex^ ) − ex^ (x + 1 ) ( 1 + ex^ )^2

N

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3. (10 Points) Evaluate the following limits. Put your answers in the boxes. Show your work.

(i) (5 points) lim x→ 1

x^4 − 1 x^2 − 1

Solution. We can factor the numerator:

lim x→ 1

x^4 − 1 x^2 − 1

= lim x→ 1

(x^2 − 1 )(x^2 + 1 ) x^2 − 1

= lim x→ 1

(x^2 + 1 ) = 12 + 1 = 2

Alternatively, we could use L’H ˆopital’s Rule:

lim x→ 1

x^4 − 1 x^2 − 1

=^ H lim x→ 1

4 x^3 2 x

= lim x→ 1

2 x^2 = 2

N

(ii) (5 points) lim x→ 0

(cos x)1/x

2

e−1/

Solution. Let the limit be L, if it exists. Then

ln L = lim x→ 0

ln(cos x)1/x

2 = lim x→ 0

ln(cos x) x^2

This limit is of the form

and so we can try L’H ˆopital’s Rule:

lim x→ 0

ln(cos x) x^2

H = lim x→ 0

1 cos x ·^ (−^ sin^ x) 2 x

= lim x→ 0

− tan x 2 x

H = lim x→ 0

− sec^2 x 2

So L = e−1/2. N

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4. (15 Points) Let f (x) = x^4 + 4 x^3 − 2. Explain your answers on each of these parts.

(i) (4 points) The derivative of f is f ′(x) = 4 x^3 + 12 x^2. On which intervals is f increasing? decreasing?

Solution. We have f ′(x) = 4 x^3 + 12 x^2 = 4 x^2 (x + 3 ). This is zero when x = 0 or x = −3. On the interval (−∞, − 3 ), f ′(x) < 0, so f is decreasing. On the intervals (−3, 0) and 0, ∞, f ′(x) > 0. So f is deccreasing on (−∞, − 3 ] and increasing on [−3, ∞). N

(ii) (4 points) The second derivative of f is f ′′(x) = 12 x^2 + 24 x. On which intervals is f concave up? concave down?

Solution. We have f ′′(x) = 12 x^2 + 24 x = 12 x(x + 2 ). This is zero when x = 0 or x = −2. On the interval (−∞, − 2 ), f ′′(x) > 0, so f is concave up. On the interval (−2, 0), f ′′(x) < 0, so f is concave down. On the interval (0, ∞), f ′′(x) > 0, so f is concave up. Thus f is concave up on (−∞, − 2 ] and [0, ∞), and concave down on [−2, 0]. N

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5. (12 Points) A department store is fencing off part of the store for children to meet and be photographed with Santa Claus. They have decided to fence off a rectangular region of fixed area 800 ft^2. Fire regulations require that there be three gaps in the fencing: 6 ft openings on the two facing sides and a 10 ft opening on the remaining wall (the fourth side of the rectangle will be against the building wall). Find the dimensions that will minimize the length of fencing used.

store wall

10 ft gap

6 ft gap 6 ft gap

fencing

Dimensions: 40 ft × 20 ft

Solution. Let x be the side with the 10 ft gap and y the side with the 6 ft gap. We want to minimize f = (x − 10 ) + 2 (y − 6 ) = x + 2 y − 2

subject to the constraint that xy = 800. Isolating y =

x

gives us a function

f (x) = x + 2 ·

x

− 2 = x +

x

The domain of this function is [10, ∞), because we need at least 10 ft to have a gap. To find the critical points, we have

f ′(x) = 1 −

x^2

So f ′(x) = 0 when x = 40 (we discard the negative root since it’s not in our domain).

Now f ′′(x) =

x^3

, which is always positive on our domain. So the unique critical point

is the global minimum of f. Thus the dimensions of the rectangle that minimize fence are 40 ft × 20 ft. N

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6. (12 Points) A cannonball is shot into the air. Its velocity is given as a function f (t) m/s, where t measured in seconds since 1:00PM. We know that f (t) takes the following values:

t 0 7.5 15 22.5 30 37.5 45 52.5 60 f (t) 10.0 6.46 5.00 3.88 2.93 2.09 1.34 0.646 0

(i) (2 points) What does the integral I =

∫ (^60)

0

f (t) dt represent?

Solution. The integral represents the distance traveled between 1:00PM and 1:01PM. N

For the two parts below, let Ln be the Riemann sum for I using n subintervals and left endpoints, Rn be the Riemann sum for I using n subintervals and right endpoints, and Mn be the Riemann sum for I using n subintervals and midpoints.

(ii) (2 points) Write out the terms in M 4 .You may leave your answer unsimplified.

Solution. We have

M 4 = 6.46 · 15 + 3.88 · 15 + 2.09 · 15 + 0.646 · 15

N

(iii) (2 points) Assume that f (t) is decreasing for all t ≥ 0. Without computing, put these in order from least to greatest: L 2 , R 4 , R 2 , I, L 4. Put your answers in the boxes. No justification is necessary. Hint. A picture might help your thinking here.

R 2 <^ R 4 <^ I <^ L 4 <^ L 2

(iv) (6 points). It turns out f (t) = 10 −

t. Compute I exactly. 200

Solution. We have

I =

[

10 t −

t3/

] 60

0

N

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8. (10 Points) Let

F(y) =

∫ (^) y

0

e−s

2 ds

y

F(y)

(i) (5 points) In each of these, select “>” if the quantity in column A is greater, “<” if the quantity in column B is greater, “=” if the two quantities are the same, and “?” if it is impossible to determine which is greater. No justification is necessary. No partial credit will be given. Please fill in the circle completely. Hint. It is mathematically impossible to compute F(y) exactly by antidifferentiation, so please do not try. That is not the point of this problem.

A B Your answer

1) F( 0 ) 0 >^ <^ =^?

2) F( 1 ) F( 2 ) >^ <^ =^?

3) F(− 1 ) F(− 2 ) >^ <^ =^?

4) F′( 0 ) 0 >^ <^ =^?

5) F′( 1 ) F′(− 1 ) >^ <^ =^?

(ii) (5 points) Suppose y(t) = 9 sin( π t) and let g(t) = F(y(t)). In other words,

g(t) =

∫ (^) 9 sin π t

0

e−s

2 ds

Find g′^

1 2

. Put your answer in the box. 0

Solution. We have g(t) = F(9 sin π t), so

g′(t) = F′(9 sin π t) · 9 π cos( π t) = 9 π e−81 sin

(^2) π t cos( π t).

Therefore g′^

1 2

= 9 π e−81 sin

(^2) ( π /2) cos

( (^) π 2

N

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9. (5 Points) Evaluate the following. No justification is necessary for this problem. In the first three, express your answer as an integer or a fraction.

(i) log 3 ( 27 ) 3

(ii) log 4

1 2

(iii) ln ( 1 ) 0

In the next two, express your answer as an angle in radians.

Note. Remember that arcsin is the inverse of sin, sometimes also written as sin−^1. But this

is not the same as

sin

(iv) arcsin

1 2

(v) arctan( 1 ) π / 4