For the scalar function $T={e}^{-r/5}\mathrm{cos}\varphi $ , determine its directional derivative along the radial direction $\hat{r}$ and then evaluate it at $P(2,\pi /4.3)$.

Ethen Blackwell
2022-07-25
Answered

For the scalar function $T={e}^{-r/5}\mathrm{cos}\varphi $ , determine its directional derivative along the radial direction $\hat{r}$ and then evaluate it at $P(2,\pi /4.3)$.

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Rihanna Robles

Answered 2022-07-26
Author has **18** answers

$\u25bd({e}^{-\frac{r}{5}}\mathrm{cos}(\varphi ))=(\frac{\frac{-{e}^{\frac{r}{5}}\mathrm{cos}(\varphi )}{-{e}^{\frac{r}{5}}\mathrm{sin}(\varphi )}}{\frac{r}{0}})$. Evaluated at $P(2,\pi /4,3)=(-1/5,-1/2.0)1/(\sqrt{2}{e}^{\frac{2}{5}})$

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Use the given graph to estimate the value of each derivative.(Round all answers to one decimal place.)Graph uploaded below.

(a) f ' (0)1

(b) f ' (1)2

(c) f ' (2)3

(d) f ' (3)4

(e) f ' (4)5

(f) f ' (5)6

(a) f ' (0)1

(b) f ' (1)2

(c) f ' (2)3

(d) f ' (3)4

(e) f ' (4)5

(f) f ' (5)6

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I find the indeterminate form of 0 or $\frac{0}{0}$. The latter tells me that L'Hopital's is an option, but since we haven't seen derivatives yet I'm not allowed to used it.

Previously I already tried swapping the $\mathrm{csc}\pi x$ for $\frac{1}{\mathrm{sin}\pi x}$ but when doing this I can't seem to get rid of the sinus. I also believe that since the limit goes to 3 and not to 0, the $\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{ax}=1$ rule is not an option either.

I tried making the $\mathrm{sin}(\frac{\pi}{2})$ so that =1, but without any success. Can anyone give me a hint?

I find the indeterminate form of 0 or $\frac{0}{0}$. The latter tells me that L'Hopital's is an option, but since we haven't seen derivatives yet I'm not allowed to used it.

Previously I already tried swapping the $\mathrm{csc}\pi x$ for $\frac{1}{\mathrm{sin}\pi x}$ but when doing this I can't seem to get rid of the sinus. I also believe that since the limit goes to 3 and not to 0, the $\underset{x\to 0}{lim}\frac{\mathrm{sin}ax}{ax}=1$ rule is not an option either.

I tried making the $\mathrm{sin}(\frac{\pi}{2})$ so that =1, but without any success. Can anyone give me a hint?

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