If a snowball melts so the its surface ar area decreases in ~ a price of $3~\frac\textcm^2\textmin$, find the rate at i beg your pardon the diameter decreases as soon as the diameter is $8$ cm.

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My price is wrong. Have the right to some one help me solve?

Given: The rate of to decrease of the surface ar area is $3~\frac\textcm^2\textmin$. I f we let $t$ it is in time in (in minutes) and also $s$ be the surface ar area (in cm$^2$), then us are provided that $\fracdsdt = -3~\textcm^3$

Unknown: The price of diminish of the diameter as soon as the diameter is $8$ centimeter

If we let $x$ be the diameter, then we want to discover $\fracdxdt$ as soon as $x = 8$ cm. If the radius is $r$ and the diameter is $x = 2r$ climate $r = \frac12x$ and also $S = 4\pi r^2$, $x=2r$, then $r=\frac12x$$S=\pi x^2 =-\frac32\pi 8$

My answer is wrong. Have the right to some one help me understand?

edited Mar 16 "15 in ~ 11:51

N. F. Taussig
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request Mar 16 "15 in ~ 11:14

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2 answers 2

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We have that the surface ar area formula is:

$$SA = 4\pi r^2$$

If us take the derivative v respect with time, we get:

$$\fracdSAdt = 8\pi r \fracdrdt$$

Now, from the trouble we are given that $\fracdSAdt = -3$, and also $r = 4$

Now us can discover $\fracdrdt$:

$$ 32\pi \fracdrdt = -3,\ \fracdrdt = -\frac332\pi$$

Now, we can also make the relation that:

$$d = 2r,\ \fracdddt = 2\fracdrdt = 2*\left(-\frac332\pi\right) = -\frac632\pi$$

Therefore ours diameter decreases at a price of $\frac632\pi$

answer Mar 16 "15 at 11:22

Varun IyerVarun Iyer
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Since the question is asking about the diameter quite than the radius, girlfriend can additionally deal directly with the diameter indigenous the beginning and also save some job-related later. This is based on the truth that $r=\fracd2$ (radius is always half the diameter).

$$SA=4 \pi r^2$$$$SA=4 \pi \bigg( \fracd2 \bigg)^2$$$$SA=4 \pi \bigg( \fracd^24 \bigg)$$$$SA=\pi d^2$$

Now you can take the derivative v respect to time.

$$\fracdSAdt=2 \pi d \fracdddt$$

Now simply plug in the given information and also solve because that $\fracdddt$.

$$-3=2 \pi (8) \cdot \fracdddt$$$$\frac-316 \pi=\fracdddt$$

So the diameter is decreasing at a rate of $\frac316 \pi$.

Here"s a an excellent website that explains this problem in a bit much more detail. The numbers are different, however the procedure is identical.


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edited january 30 "19 at 6:35
answered jan 8 "19 in ~ 15:02

Jake OJake O
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