Math 101 A04 Midterm 1 Solutions and grading guide

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Math 101 A04. Midterm 1. Solutions and grading guide. 1. (7 points) Suppose there is a metal rod that is 10 cm long. As you move along the rod. (x goes from 0  ...
Math 101 A04

Midterm 1

Solutions and grading guide

1. (7 points) Suppose there is a metal rod that is 10 cm long. As you move along the rod (x goes from 0 to 10), the linear density at x is d(x) = 5 + x2 grams per centimeter. (a) (3 points) Break the rod into 5 equal-sized segments. Write a Riemann sum approximating the mass of the rod by assuming the density is nearly constant on each segment (use the left endpoint of each interval to evaluate the density function).

Breaking the rod into 5 equal pieces means making each piece of length 2. The left endpoints of these subintervals are x1 = 0, x2 = 2, x3 = 4, x4 = 6, x4 = 8. P So the Riemann sum is 5i=1 (5 + x2i ) · 2 = 5 · 2 + 9 · 2 + 21 · 2 + 41 · 2 + 69 · 2 Grading: 1 point for making clear somehow what the xi ’s are, 1 point for ∆x = 2 or using n = 5 (knowing there are five pieces), 1 point for complete answer. (b) (2 points) As the number of segments goes to infinity, the Riemann sums approach a certain integral. What is the integral?

Z

10

5 + x2 dx 0

Grading: A common point of confusion was not getting the correct endpoints for the interval of integration. (c) (2 points) Evaluate the integral from the previous part. This gives the total mass of the rod.

5x +

x3 10 103 ]0 = 5 · 10 + 3 3

2. (7 points) Consider the region R bound by the two curves y = x2 and y = 4x. Revolve it around the line x = 6 to obtain a solid. Find the volume of the solid by using the disc/washer method. [Hint: start by graphing the curves and noting the points of intersection.]

The two curves intersect at (0, 0) and (4, 16) (just solve by setting x2 = 4x). Between the two points, y = 4x is to the left of y = x2 for a given y (draw a picture!). To decompose the solid into washers, you slice the region R into slices of thickness dy. Each slice is revolved around x = 6 to become a washer shape. The area of a washer is π(R2 − r 2 ) where R is the outer radius and r is the inner radius. Since we will have dy in our integral, we need everything in terms of y, which means R and r needs to be in terms of y. The picture should make it clear that R is 6 − the x-coordinate of point on y = 4x. This x coordinate in terms of y is x = 41 y. So R = 6 − 41 y. √ Similarly, you should be able to obtain r = 6 − y.

So the volume integral is Z 16 Z 16 √ 1 2 2 π[R − r ] dy = π[(6 − y)2 − (6 − y)2 ] dy. 4 0 0 16  Z 16 √ 1 2 1 3 3/2 2 = y + 12 y − 4y dy = π y + 8y − 2y Evaluating the integral gives π 16 48 0 0   2 16 + 8 · 43 − 2 · 162 . π 3 Grading: 1 point for figuring out points of intersection, 2 point for basic washer formula in integral with correct dy and limits of integration, 2 points for radii formula, 2 points for the rest.

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3. (4 points) Consider the same solid from the previous problem. Set up an integral for the volume using the cylindrical shell method. Do not evaluate the integral.

Recall that the shell is obtained by taking a vertical slice of the region with infinitesimal thickness dx and rotating it around x = 6. Z 4 We know the integral is of the form 2π · r · h dx where r and h are functions of x. h 0

is the height of a cylindrical shell and r is the radius of the shell.

Looking at your picture from the previous problem, it should be clear that the radius is 6 − x and the height is 4x − x2 . So the volume of a shell of thickness dx is 2π(6 − Z 4 2 x)(4x − x )dx. So the volume integral is 2π(6 − x)(4x − x2 ) dx. 0

Grading: 1 point for limits of integration and dx in integral, 1 point for knowing something like 2πrh dx for cylindrical shell, 1 point for r, 1 point for h

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4. (5 points) Find the arc length of the curve y = x2 −

1 ln x from x = 1 to x = 2. 8

The method is to integrate the arc length q element ds. Since the curve is parametrized dy 2 ) dx. by x, it is natural to use the form of ds = 1 + ( dx dy dx

= 2x −

1 8x

dy 2 and ( dx ) = 4x2 − 12 +

dy 2 ) = 4x2 + 21 + 1 + ( dx

1 64x2

= (2x +

1 . 64x2

1 2 ) . 8x

1 So ds = (2x + 8x ) dx and the arc length is 2 Z Z 2 1 1 1 2 dx = x + ln x = 3 + ln 2 ds = 2x + 8x 8 8 1 1

Grading: 1 point for knowing the formula for ds, 1 point for computing derivative dy/dx, 2 points for knowing how to simplify ds to eliminate radical, 1 point for the rest

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5. (5 points) Take the curve from the previous problem and rotate it around the y-axis to 1 obtain a surface. What is the surface area? [The curve was y = x2 − ln x from x = 1 8 to x = 2]

The basic form of the surface integral looks like Z b 2πr ds, where ds is an infinitesimal bit of the curve and r is the distance from the a

axis of revolution to the bit of curve.

Note that we already worked out ds in terms of x in the previous problem and that r = x, so it’s natural to set up the integral in terms of x.  3 2 Z 2 Z 2 1 1 1 2x 2 2πx(2x + ) dx = 2π + x 2x + dx = 2π 8x 8 3 8 1 1 1 Grading: 2 points for the general form of the integral with apppropriate ds formula, 1 point for r, 1 point for setting up the integral for antidifferentiation, 1 point for antidifferentiation

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6. (12 points) Find the derivatives of the following functions: √ 1−x2 (a) 3 (b) (sin−1 x2 )3 (c) sec−1 2x (a) Recall the derivative of ax is ax · ln a. √ √ d √1−x2 d√ 1 2 2 ) = 3 1−x ln 3 1 − x2 = 3 1−x ln 3 (1 − x2 )−1/2 (−2x) So (3 dx dx 2 1 1 d 2 d (sin−1 x2 ) = 3(sin−1 x2 )2 √ (x ) = 3(sin−1 x2 )2 √ (2x) 4 dx 1 − x dx 1 − x4 1 √ (c) Recall the derivative of sec−1 x is . So by the chain rule, the answer is |x| x2 − 1 1 p · 2. |2x| (2x)2 − 1 (b) 3(sin−1 x2 )2

Grading: 4 points for each part. If you knew the basic derivative for au or sin−1 u, that was 1 point each. 2 points for derivative of sec−1 x with 1 point for absolute value sign. The rest of the points were for using the chain rule and substitution correctly.

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7. (14 points) Find the antiderivatives of the following functions: 1 (a) (4 points) 1 + 9x2 x (b) (5 points) √ 2x + 1 (c) (5 points) x cosh x (a) This looks suspiciously like the derivative of inverse tangent. So we make the substitution u = 3x. ZThen du = 3dx and Z 1 1 1 1 −1 dx = du = tan u = tan−1 3x 1 + 9x2 3(1 + u2 ) 3 3 Grading: 2 points for u and du. 1 point for derivative of tan−1 x. 1 point for complete answer. (b) There are two ways you could have used. Method 1: We use a u-substitution. Z u−1 x 1 √ √ du = Set u = 2x+ 1. Then du = 2 dx. Note x = Then dx = 4  u 2x + 1    Z 1 2 1 2 1 u1/2 − u−1/2 du = u3/2 − 2u1/2 = (2x + 1)3/2 − 2(2x + 1)1/2 . 4 4 3 4 3 √ There’s also a way using u = 2x + 1. u−1 . 2

Z

Method 2: We use integration by parts. √ Then du = dx and v = 2x + 1. R R√ √ √ So the antiderivative is uv− vdu = x 2x + 1− 2x + 1 dx = x 2x + 1− 31 (2x+1)3/2 .

Set u = x and dv =

√ 1 dx. 2x+1

Both methods, of course, give the same answer, but there is some algebraic rearranging that has to be done first before it is obvious. Grading: Only several people used Method 1 and I think only one did it correctly. For Method 2, 1 point for u and dv. 2 points for du and v. Then 1 point for using the correct integration by parts formula, and 1 point for correct completion of problem. (c) Again integration by parts. Set u = x and dv = cosh xdx. Then du = dx and v = sinh x. R R So uv − vdu = x sinh x − sinh xdx = x sinh x − cosh x.

In both cases of integration by parts, u = x seems a natural choice because taking R the derivative simplifies x to just 1 while we know how to antidifferentiate the dv. So vdu should hopefully be do-able, which it is. Grading: 1 point for u and dv. 2 points for du and v. 1 point for correct integration by parts formula, and 1 point for completing correctly.

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