Question

Help calculus module 6 DBQ please show work

Answer 1

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 $\boxed{0\le t$.