Exam B Solutions

  1. Find the derivative of f(x)=sin⁡2x1+x2f(x)=\frac{\sin ^{2} x}{\sqrt{1+x^{2}}}. Do not simplify your answer.

    Solution: Applying the quotient rule, we obtain f′(x)=1+x2ddx[(sinx)2]−sin⁡2x⋅ddx[(1+x2)1/2]1+x2=1+x2⋅2(sin⁡x)(cos⁡x)−sin⁡2x⋅12(1+x2)−1/2(2x)1+x2\begin{aligned} f^{\prime}(x) & =\frac{\sqrt{1+x^{2}} \frac{d}{d x}\left[(\sin x)^{2}\right]-\sin ^{2} x \cdot \frac{d}{d x}\left[\left(1+x^{2}\right)^{1 / 2}\right]}{1+x^{2}} \\ & =\frac{\sqrt{1+x^{2}} \cdot 2(\sin x)(\cos x)-\sin ^{2} x \cdot \frac{1}{2}\left(1+x^{2}\right)^{-1 / 2}(2 x)}{1+x^{2}} \end{aligned}

  2. If f(9)=1f(9)=1, f′(9)=3f^{\prime}(9)=3, g(1)=0g(1)=0, g′(1)=−2g^{\prime}(1)=-2, and h(x)=x3/2eg(f(x))h(x)=x^{3 / 2} e^{g(f(x))}, then what is h′(9)h^{\prime}(9)? Simplify your answer.

    Solution: Applying the product rule, we have h′(x)=x3/2ddx[eg(f(x))]+eg(f(x))ddx[x3/2]=x3/2eg(f(x))g′(f(x))f′(x)+eg(f(x))(3/2)x1/2\begin{aligned} h^{\prime}(x) & =x^{3 / 2} \frac{d}{d x}\left[e^{g(f(x))}\right]+e^{g(f(x))} \frac{d}{d x}\left[x^{3 / 2}\right] \\ & =x^{3 / 2} e^{g(f(x))} g^{\prime}(f(x)) f^{\prime}(x)+e^{g(f(x))}(3 / 2) x^{1 / 2} \end{aligned} Thus, h′(9)=93/2eg(1)g′(1)f′(9)+eg(1)(3/2)91/2=27e0(−2)(3)+e0(3/2)(3)=−3152h^{\prime}(9)=9^{3 / 2} e^{g(1)} g^{\prime}(1) f^{\prime}(9)+e^{g(1)}(3 / 2) 9^{1 / 2}=27 e^{0}(-2)(3)+e^{0}(3 / 2)(3)=-\frac{315}{2}

  3. Below is a portion of the graph of a function ff.

    Graph of y = f(x) on approximately x = -6 to x = 5. The function has open circles at x = -2 and x = 1 indicating these points are not in the domain. At x = -4 the function has a sharp corner. At x = -2 there is a jump discontinuity with different left and right limits. At x = 1 the point is removed but the limit exists. At x = 3 there is a discontinuity where the limit does not exist. The curve is labeled y = f(x).
    Original (raster)
    Graph of y = f(x) on approximately x = -6 to x = 5. The function has open circles at x = -2 and x = 1 indicating these points are not in the domain. At x = -4 the function has a sharp corner. At x = -2 there is a jump discontinuity with different left and right limits. At x = 1 the point is removed but the limit exists. At x = 3 there is a discontinuity where the limit does not exist. The curve is labeled y = f(x). (TikZ reconstruction)
    Reconstructed (TikZ) — RMSE: 0.1616 FAIL

    For the following, give all values of aa in the interval (−5,4)(-5,4) satisfying the given condition. If there are none, write "none". No work is required.

    1. all aa not in the domain of ff: −2,1-2, \quad 1

    2. all aa such that ff is not continuous at aa: −2,1,3-2, \quad 1, \quad 3

    3. all aa such that lim⁡x→af(x)\lim_{x \rightarrow a} f(x) does not exist: −2,3-2, \quad 3

    4. all aa at which ff is not differentiable: −4,−2,1,3-4, \quad -2, \quad 1, \quad 3

    Grade the problem “correct” even if you made one careless error in filling the blanks.

  4. Find and classify the critical numbers of f(x)=x4/5(x−4)2f(x)=x^{4 / 5}(x-4)^{2}, indicating for each critical number whether it yields a relative maximum value of ff, a relative minimum value, or neither.

    Solution: Applying the product rule, we obtain f′(x)=x4/52(x−4)+(x−4)2(4/5)x−1/5=25x−1/5(x−4)(5x+2(x−4))=2x−1/5(x−4)(7x−8).f^{\prime}(x)=x^{4 / 5} 2(x-4)+(x-4)^{2}(4 / 5) x^{-1 / 5}=\frac{2}{5} x^{-1 / 5}(x-4)(5 x+2(x-4))=2 x^{-1 / 5}(x-4)(7 x-8) . Thus, 00, 44 and 8/78/7 are the critical numbers of ff. To classify these critical numbers, we determine the signs of f′f^{\prime}:

    Number line sign chart for f'(x). Critical numbers 0, 8/7, and 4 are marked. The sign of f' is negative for x < 0, positive for 0 < x < 8/7, negative for 8/7 < x < 4, and positive for x > 4.
    Original (raster)
    Number line sign chart for f'(x). Critical numbers 0, 8/7, and 4 are marked. The sign of f' is negative for x < 0, positive for 0 < x < 8/7, negative for 8/7 < x < 4, and positive for x > 4. (TikZ reconstruction)
    Reconstructed (TikZ) — RMSE: 0.2265 FAIL

    Applying the first derivative test (note ff is continuous at each of its critical numbers), we see that ff has a local minimum at 00, a local maximum at 8/78/7, and a local minimum at 44.

  5. Find the area of the region bounded by the curves y=x2y=x^{2} and y=x+2y=x+2.

    Solution: The region bounded by the curves y=x2y=x^{2} and y=x+2y=x+2 is shaded blue in the figure below.

    Graph showing the region bounded by the parabola y = x squared and the line y = x + 2. The curves intersect at x = -1 and x = 2. The region between them is shaded blue, with the line above and the parabola below.
    Original (raster)
    Graph showing the region bounded by the parabola y = x squared and the line y = x + 2. The curves intersect at x = -1 and x = 2. The region between them is shaded blue, with the line above and the parabola below. (TikZ reconstruction)
    Reconstructed (TikZ) — RMSE: 0.1153 PASS

    Its area is given by ∫−12(x+2−x2)dx=[x22+2x−x33]−12=(2+4−8/3)−(1/2−2+1/3)=92\int_{-1}^{2}\left(x+2-x^{2}\right) d x=\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{2}=(2+4-8 / 3)-(1 / 2-2+1 / 3)=\frac{9}{2}

  6. A canister is dropped from a helicopter 500 m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact speed of 120m/s120 \mathrm{~m} / \mathrm{s}. Will it burst? Assume that after it is dropped, the canister experiences (only) the acceleration due to gravity, approximately 10m/s210 \mathrm{~m} / \mathrm{s}^{2}.

    Solution: Let h(t)h(t) be the height of the canister, in meters, tt seconds after it’s dropped (until it hits the ground). We have h′′(t)=−10,h(0)=500,andh′(0)=0(because the canister is dropped).h^{\prime \prime}(t)=-10, \quad h(0)=500, \quad \text{and} \quad h^{\prime}(0)=0 \quad \text{(because the canister is dropped).} Integrating both sides of h′′(t)=−10h^{\prime \prime}(t)=-10 yields h′(t)=−10t+Ch^{\prime}(t)=-10 t+C, and h′(0)=0h^{\prime}(0)=0 shows C=0C=0. Thus, h′(t)=−10th^{\prime}(t)=-10 t. Integrating both sides of h′(t)=−10th^{\prime}(t)=-10 t and using h(0)=500h(0)=500, we find h(t)=−5t2+500h(t)=-5 t^{2}+500. The canister hits the ground when h(t)=0h(t)=0; that is, when −5t2+500=0-5 t^{2}+500=0, so that t=±10t= \pm 10, but t=−10t=-10 is irrelevant. Thus, the canister hits the ground after 10 seconds. Its velocity at that instant is h′(10)=−100m/sech^{\prime}(10)=-100 \mathrm{~m} / \mathrm{sec}. Its impact speed is 100m/sec100 \mathrm{~m} / \mathrm{sec}; so, we conclude that the canister should not burst.

    Remark 1. When grading this problem, consider your solution “correct” provided you found that the canister’s impact speed is 100m/s100 \mathrm{~m} / \mathrm{s}.

  7. Find ∫ex1+exdx\int \frac{e^{x}}{1+e^{x}} d x

    Solution: ∫ex1+exdx=∫1udu(u=1+ex;du=exdx)=ln⁡|u|+C=ln⁡|ex+1|+C=ln⁡(ex+1)+C\begin{aligned} \int \frac{e^{x}}{1+e^{x}} d x & =\int \frac{1}{u} d u \quad\left(u=1+e^{x} ; d u=e^{x} d x\right) \\ & =\ln |u|+C \\ & =\ln \left|e^{x}+1\right|+C \\ & =\ln \left(e^{x}+1\right)+C \end{aligned} Grade the problem “correct” even if you forgot to include +C+C.

  8. Evaluate ∫0π/12tan⁡3(3x)sec⁡2(3x)dx\int_{0}^{\pi / 12} \tan ^{3}(3 x) \sec ^{2}(3 x) d x

    Solution: ∫0π/12tan⁡3(3x)sec⁡2(3x)dx=13∫01u3du(u=tan(3x),du=3sec⁡2(3x)dx,13du=sec⁡2(3x)dx)=13[u44]01=112\begin{aligned} \int_{0}^{\pi / 12} \tan ^{3}(3 x) \sec ^{2}(3 x) d x & =\frac{1}{3} \int_{0}^{1} u^{3} d u \quad\left(u=\tan (3 x), d u=3 \sec ^{2}(3 x) d x, \frac{1}{3} d u=\sec ^{2}(3 x) d x\right) \\ & =\frac{1}{3}\left[\frac{u^{4}}{4}\right]_{0}^{1} \\ & =\frac{1}{12} \end{aligned}

  9. Over the time interval 1≤t≤1001 \leq t \leq 100 the temperature of a freezer compartment is given by f(t)=t−4t2f(t)=\frac{t-4}{t^{2}} where tt is measured in hours and f(t)f(t) is measured in degrees Celsius. What’s the maximum temperature of the freezer over this time interval?

    Solution: Because ff is continuous on [1,100][1,100] (it’s a rational function whose domain includes the entire interval [1,100][1,100]), the Extreme-Value Theorem tells us that ff attains a maximum value at some number, say cc, in [1,100][1,100]. This number cc must either be an endpoint of [1,100][1,100] or a critical number of ff in (1,100)(1,100). We have f′(t)=t2(1)−(t−4)2tt4=8−tt3f^{\prime}(t)=\frac{t^{2}(1)-(t-4) 2 t}{t^{4}}=\frac{8-t}{t^{3}} and thus 88 is the only critical number of ff inside (1,100)(1,100). The largest of the three numbers f(1),f(8),f(100)f(1), f(8), f(100) will be the maximum value of ff on [1,100][1,100]. Since f(1)=−3,f(8)=1/16f(1)=-3, f(8)=1 / 16, and f(100)=96/10000<100/10000=1/100f(100)=96 / 10000< 100 / 10000=1 / 100, we see the maximum temperature of the freezer over the time interval [1,100][1,100] is 1/161/16 degrees Celsius.

  10. Both Rugby Road and Madison Lane meet University Avenue at right angles. (See the diagram below.) Dolly is biking on University Ave at 9ft/sec9 \mathrm{ft} / \mathrm{sec} and has passed Madison Ln heading toward Rugby. James is biking away from University Ave along Rugby traveling at 8ft/sec8 \mathrm{ft} / \mathrm{sec}. At what rate is the distance between Dolly and James changing at the instant when James is 40 ft from and Dolly is 30 ft from the intersection of Rugby Rd and University Ave? Is the distance between them increasing or decreasing at this instant?

    Diagram showing the intersection of Rugby Road, Madison Lane, and University Avenue at right angles. Rugby Road extends vertically upward on the left, Madison Lane extends vertically upward on the right, and University Avenue runs horizontally at the bottom. The variable y marks the distance along Rugby Road from the intersection, x marks the distance along University Avenue, and z marks the diagonal distance between Dolly on University Ave and James on Rugby Rd.
    Original (raster)
    Diagram showing the intersection of Rugby Road, Madison Lane, and University Avenue at right angles. Rugby Road extends vertically upward on the left, Madison Lane extends vertically upward on the right, and University Avenue runs horizontally at the bottom. The variable y marks the distance along Rugby Road from the intersection, x marks the distance along University Avenue, and z marks the diagonal distance between Dolly on University Ave and James on Rugby Rd. (TikZ reconstruction)
    Reconstructed (TikZ) — RMSE: 0.2003 FAIL

    As suggested by the diagram above, we let xx be the distance, in feet, from Dolly to the intersection of Rugby and University and yy be the distance, in feet, from James to the intersection. Finally, we let zz be the distance, in feet, from Dolly to James. We are given that dxdt=−9ft/sec and dydt=8ft/sec.\frac{d x}{d t}=-9 \mathrm{ft} / \mathrm{sec} \quad \text { and } \quad \frac{d y}{d t}=8 \mathrm{ft} / \mathrm{sec} . We seek dzdt\frac{d z}{d t} when x=30,y=40x=30, y=40, and, by the Pythagorean Theorem, z=50z=50.
    The Pythagorean Theorem yields, more generally, that x2+y2=z2.x^{2}+y^{2}=z^{2}. Differentiating the preceding equation with respect to time, we obtain 2xdxdt+2ydydt=2zdzdt,2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 z \frac{d z}{d t}, which yields dzdt=1z(xdxdt+ydydt).\frac{d z}{d t}=\frac{1}{z}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right). At the time when x=30x=30 and y=40y=40, we have dzdt=150(30(−9)+40(8))=1ft/sec.\frac{d z}{d t}=\frac{1}{50}(30(-9)+40(8))=1 \mathrm{~ft} / \mathrm{sec}. Thus when x=30x=30 and y=40y=40 the distance between Dolly and James is increasing at the rate of 1 ft/sec.