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damaskus [11]
3 years ago
10

Help calculus module 6 DBQ please show work

Mathematics
1 answer:
Sloan [31]3 years ago
6 0

1. Let a,b,c be the three points of intersection, i.e. the solutions to f(x)=g(x). They are approximately

a\approx-3.638

b\approx-1.862

c\approx0.889

Then the area R+S is

\displaystyle\int_a^c|f(x)-g(x)|\,\mathrm dx=\int_a^b(g(x)-f(x))\,\mathrm dx+\int_b^c(f(x)-g(x))\,\mathrm dx

since over the interval [a,b] we have g(x)\ge f(x), and over the interval [b,c] we have g(x)\le f(x).

\displaystyle\int_a^b\left(\dfrac{x+1}3-\cos x\right)\,\mathrm dx+\int_b^c\left(\cos x-\dfrac{x+1}3\right)\,\mathrm dx\approx\boxed{1.662}

2. Using the washer method, we generate washers with inner radius r_{\rm in}(x)=2-\max\{f(x),g(x)\} and outer radius r_{\rm out}(x)=2-\min\{f(x),g(x)\}. Each washer has volume \pi({r_{\rm out}(x)}^2-{r_{\rm in}(x)}^2), so that the volume is given by the integral

\displaystyle\pi\int_a^b\left((2-\cos x)^2-\left(2-\frac{x+1}3\right)^2\right)\,\mathrm dx+\pi\int_b^c\left(\left(2-\frac{x+1}3\right)^2-(2-\cos x)^2\right)\,\mathrm dx\approx\boxed{18.900}

3. Each semicircular cross section has diameter g(x)-f(x). The area of a semicircle with diameter d is \dfrac{\pi d^2}8, so the volume is

\displaystyle\frac\pi8\int_a^b\left(\frac{x+1}3-\cos x\right)^2\,\mathrm dx\approx\boxed{0.043}

4. f(x)=\cos x is continuous and differentiable everywhere, so the the mean value theorem applies. We have

f'(x)=-\sin x

and by the MVT there is at least one c\in(0,\pi) such that

-\sin c=\dfrac{\cos\pi-\cos0}{\pi-0}

\implies\sin c=\dfrac2\pi

\implies c=\sin^{-1}\dfrac2\pi+2n\pi

for integers n, but only one solution falls in the interval [0,\pi] when n=0, giving c=\sin^{-1}\dfrac2\pi\approx\boxed{0.690}

5. Take the derivative of the velocity function:

v'(t)=2t-9

We have v'(t)=0 when t=\dfrac92=4.5. For 0\le t, we see that v'(t), while for 4.5, we see that v'(t)>0. So the particle is speeding up on the interval \boxed{\dfrac92 and slowing down on the interval \boxed{0\le t.

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