MA005: Calculus I

Practice 1: The surface area of the cylinder is $\mathrm{SA}=2 \pi \mathrm{rh}+2 \pi \mathrm{r}^{2}$. From the Example, we know that $\mathrm{h'}$  $=7$ $\mathrm{m} / \mathrm{s}$ and $\mathrm{r}^{\prime}=3 \mathrm{~m} / \mathrm{s}$, and we want to know how fast the surface area is changing when $\mathrm{h}=5 \mathrm{~m}$ and $\mathrm{r}=6 \mathrm{~m}$.

Practice 2: The volume of the cylinder is $\mathrm{V}=($ area of the bottom) (height $)=\pi \mathrm{r}^{2} \mathrm{~h}$. We are told that $\mathrm{r}^{\prime}=-3$ $\mathrm{m} / \mathrm{s}$, and that $\mathrm{h}=5 \mathrm{~m}, \mathrm{r}=6 \mathrm{~m}$, and $\mathrm{h}^{\prime}=7 \mathrm{~m} / \mathrm{s}$

Practice 3: Fig. 23 represents the situation described in this problem. We are told that $\mathrm{L}^{\prime}=-30 \, \mathrm{ft} / \mathrm{min}.$ The variable $\mathrm{F}$ represents the distance of the fish from the angler, and we are asked to find $F'$, the rate of change of $F$ when $L$ $=60 \, \mathrm{ft}$.

Fortunately, the problem contains a right triangle so there is a formula (the Pythagorean formula) connecting $F$ and $L: F^{2}+10^{2}=L^{2}$ so

$\mathrm{F}=\sqrt{\mathrm{L}^{2}-100}$.

Then $F^{\prime}=\frac{1}{2}\left(L^{2}-100\right)^{-1 / 2} \frac{\mathrm{d}\left(\mathrm{L}^{2}-100\right)}{\mathrm{dt}}=\frac{2 \mathrm{~L} \cdot \mathrm{L}^{\prime}}{2 \sqrt{\mathrm{L}^{2}-100}}$.

When $L=60$ feet, $F^{\prime}=\frac{2(60 \, \mathrm{ft})(-30 \, \mathrm{ft} / \mathrm{min})}{2 \sqrt{(60 \, \mathrm{ft})^{2}-(10 \, \mathrm{ft})^{2}}} \approx \frac{-3600 \, \mathrm{ft}^{2} / \mathrm{min}}{118.32 \, \mathrm{ft}} \approx-30.43 \, \mathrm{ft} / \mathrm{min}$.

We could also find $\mathrm{F}$ ' implicitly: $\mathrm{F}^{2}=\mathrm{L}^{2}-100$ so, differentiating each side,

Then we could use the given values for $\mathrm{L}$ and $\mathrm{L}$ ' and value of $\mathrm{F}$ (found using the Pythagorean formula) evaluate $\mathrm{F'}$.