Math 231E. Fall 2013. Midterm 2 Practice Solutions.Problem 1. Compute the derivatives of the following functionsa. f(x) = cos(x2)esin(x)b. f(x) = arctan(x)c. f(x) = ln(ex(1 + x2+ x3))Solution:a. We use the product and chain rules. We haveddx(cos(x2)esin(x)) =ddxcos(x2)esin(x)+ cos(x2)ddx(esin(x))= −2x sin(x2)esin(x)+ cos(x) cos(x2)esin(x).b. If y = arctan(x), then x = tan(y). Using Implicit Differentiation, we have1 = sec2(y)dydxordydx= cos2(y) = cos2(tan−1(x)).If we draw a triangle whose interior angle θ has the property that tan(θ) = x, thenwe can make the opposite site x and the adjacent side 1, which gives a hypotenuse of√1 + x2, and thus cos2(tan−1(x)) = 1/1 + x2, sodydx=11 + x2.c. We first simplifyln(ex(1 + x2+ x3)) = ln(ex) + ln(1 + x2+ x3) = x + ln(1 + x2+ x3),and then we havef0(x) = 1 +2x + 3x21 + x2+ x3.Problem 2. Consider the polynomial P (x) = 8x3− 36x2+ 46x − 15. We want to find theroots of this polynomial. Find “brackets” for all of the roots of this polynomial of the form[k, k + 1], i.e. find three integers k1, k2, k3such that you can show there is a root r withk1< r < k1+ 1. Use any calculus facts we have covered so far in class.Solution: We plug in a few integer values. Notice thatP (0) = −15, P (1) = 3, P (2) = −3, P (3) = 15.By the Intermediate Value Theorem, we know that we must have a root in [0, 1], one in [1, 2],and one in [2, 3].Problem 3. Evaluate the following limits, or show they do not exist:a. limx→11 − x2+ 4x3sin(x)b. limx→0|x|sin(x)c. limx→0cos ex3− 1x − sin(x)!d. limx→3Zx3e−t2dt.e. limx→31x − 3Zx3e−t2dt.f. limx→0exp(x/ sin(x))Solution:a. We see that we can just plug in x = 1 and obtain 4/ sin(1).b. This limit will not exist. Using the Taylor series expansion for sin(x) = x −x3/6 +. . . ,we havelimx→0+|x|sin(x)= limx→0+xsin(x)= 1,butlimx→0−|x|sin(x)= limx→0+−xsin(x)= −1,so the limit cannot exist.c. Using the Taylor expansionsex3= 1 + x3+x62+ O(x9), x −sin(x) =x36+ O(x5),we havelimx→0ex3− 1x − sin(x)= limx→0x3+x62+ O(x9)x36+ O(x5)= 6,and since cos(x) is continuous at 6,limx→0cos ex3− 1x − sin(x)!= cos(6).d. Notice that 0 ≤ |e−t2| ≤ 1 for all t, so thatZx30 dt ≤Zx3e−t2dt ≤Zx3dx,or0 ≤Zx3e−t2dt ≤ (x −3).Sincelimx→30 = limx→3(x − 3) = 0,by the Squeeze Theorem we havelimx→3Zx3e−t2dt = 0.e. There are three techniques for this solution.• Let us writef(x) =Zx3e−t2dxWe know from before that f(3) = 0. The FTC tells us that f0(x) = e−x2, andf00(x) = −2xe−x2. This means that the Taylor series for f(x) isf(x) = f(3)+f0(3)(x−3)+f00(3)2(x−3)2+O(x−3)3= e−9(x−3)−3e−9(x−3)2+O(x−3)3.Thenlimx→3f(x)x − 3= e−9+ O(x −3) = e−9.• Define f(x) as before, and recall that f (3) = 0, so f(x) = f(x) − f(3). We thenhavelimx→3f(x) − f(3)x − 3= f0(3),and by the FTC we know f0(x) = e−x2, so f0(3) = e−9.• Finally, by the previous problem, we know thatlimx→3f(x)x − 3is an indeterminant (0/0) form, since limx→3f(x) = 0. So we can use l’Hˆopital’sRule, and obtainlimx→3f0(x)1= e−9.f. Since exis continuous,limx→0expxsin(x)= explimx→0xsin(x)= e1= e.Problem 4. Let us definef(x) = “the second digit after the decimal point in the decimal expansion for x”.Does limx→0f(x) exist? If so, what is it? If not, why not?Solution: Yes, and it is zero. To see this, notice that when a number is small, whether itis positive or negative, many digits in its decimal expansion will be zero. In particular, if|x| < 1/100, then regardless of the sign of x, the first two digits of x’s decimal expansion willbe zero.Problem 5. A parallelogram has sides of length one meter that are hinged at the ends.At the moment when the height is√22m it is decreasing at 1 m/s. How fast is the anglechanging?Solution: The height of the parallelogram is sin(θ) · 1 m. Differentiating the equation h =sin(θ) m givesdhdt= cos(θ) ·dθdtm.If the height is√2/2 m, then θ = π/4 radians, and cos(θ) =√2/2. Sodθdt= (−1 m/s)(√2) rad/m = −√2 rad/s.Problem 6. Evaluate the following definite integralsa.Zπ0x cos(x) dxb.Z21x2ln x dx c.Z0−1x(1 + x)2012dxSolution:a. We will integrate by parts, choosing u = x and v = cos(x) dx. This makes du = dxand v = sin(x), so we haveZπ0x cos(x) dx = x sin(x)x=πx=0−Zπ0sin(x) dx= x sin(x)x=πx=0+ cos(x)x=πx=0= π sin(π) −0 + cos(π) − cos(0) = −2.b. We will choose u = ln x, dv = x2dx, giving du = dx/x and v = x3/3, soZ21x2ln x dx =x33ln xx=2x=1−Z21x23dx=x33ln xx=2x=1−x39x=2x=1=83ln 2 − 0 −89+19=83ln 2 −79.c. We will choose u = x, dv = (1 + x)2012, dx, giving du = dx and v = (1 + x)2013/2013,soZ0−1x(1 + x)2012, dx = x(1 + x)20132013x=0x=−1−Z0−1(1 + x)20132013dx= x(1 + x)20132013x=0x=−1−(1 + x)2014(2013)(2014)x=0x=−1= 0 − 0 −1(2013)(2014)− 0 = −1(2013)(2014).Problem 7. Evaluate the following indefinite integralsa.Z11 +√xdx b.Z√xe√xdxc.Zsin(θ)1 − cos2(θ)dθSolution:a. We try the u-substitution u = 1 +√x, or du = dx/2√x. Solving for dx givesdx = 2√x du = 2(u − 1) du,so our integral becomesZdx1 +√x=Z2(u − 1) duu=Z2 −2udu = 2u − 2 ln |u|+ C.Plugging in for u givesZdx1 +√x= 2(1 +√x) − 2 ln |1 +√x| + C = 2√x − 2 ln |1 +√x| + C.Note that we can simplify and remove any constant we like, since it gets absorbed inthe arbitrary constant C.b. We try the substitution w =√x (we don’t use u here, because it would cause troublein a bit). This givesdw =12√xdx, dx = 2√x dw = 2w dw,soZ√xe√xdx = 2Zw2ewdw.We now integrate by parts. Choose u = w2and dv = ewdw, and then we have2Zw2ewdw = 2w2ew− 2Z2we2dw = 2w2ew− 4Zwewdw.Again integrating by parts and choosing u = w, dv = ewdw, we haveZwewdw = wew−Zewdw = wew− ew+ C.Therefore,2Zw2ewdw = 2w2ew− 4(wew− ew+ C) = 2w2ew− 4wew+ 4ew+ C,andZ√xe√xdx = 2xe√x− 4√xe√x+ 4e√x+ C.c. We make the substitution u = cos(θ), and du = −sin(θ) dθ, to obtainZsin(θ)1 − cos2(θ)dθ =Z−du1 − u2.Noting that11 + u+11 − u=21 − u2,we haveZ−du1 − u2= −12ln |1 + u| +12ln |1 − u| + C =12ln |1 − cos(θ)|−12ln |1 + cos(θ)|+ C.Problem 8. A particle at the end of a string spins in a circle about the origin. At theinstant that the particle is at the point (1/2 m,√3/2 m), the x coordinate is decreasing at arate of 1
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