You are watching: If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min find the rate
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.
edited january 30 "19 at 6:35
answered jan 8 "19 in ~ 15:02
Jake OJake O
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