If a snowball melts so that its surface area decreases at a price of $3~frac extcm^2 extmin$, find the rate at which the diameter decreases once the diameter is $8$ cm.

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My answer is wrong. Can some one aid me solve?

Given: The rate of decrease of the surchallenge location is $3~frac extcm^2 extmin$. I f we let $t$ be time in (in minutes) and also $s$ be the surchallenge area (in cm$^2$), then we are offered that $fracdsdt = -3~ extcm^3$

Unknown: The price of decrease of the diameter as soon as the diameter is $8$ cm

If we let $x$ be the diameter, then we desire to uncover $fracdxdt$ once $x = 8$ cm. If the radius is $r$ and the diameter is $x = 2r$ then $r = frac12x$ and also $S = 4pi r^2$, $x=2r$, then $r=frac12x$$S=pi x^2 =-frac32pi 8$

My answer is wrong. Can some one assist me understand?

edited Mar 16 "15 at 11:51

N. F. Taussig
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asked Mar 16 "15 at 11:14

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2 Answers 2

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We have actually that the surconfront location formula is:

$$SA = 4pi r^2$$

If we take the derivative via respect with time, we get:

$$fracdSAdt = 8pi r fracdrdt$$

Now, from the difficulty we are offered that $fracdSAdt = -3$, and $r = 4$

Now we have the right to find $fracdrdt$:

$$ 32pi fracdrdt = -3, fracdrdt = -frac332pi$$

Now, we deserve to also make the relation that:

$$d = 2r, fracdddt = 2fracdrdt = 2*left(-frac332pi ight) = -frac632pi$$

Thus our diameter decreases at a price of $frac632pi$

answered Mar 16 "15 at 11:22

Varun IyerVarun Iyer
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Due to the fact that the question is asking around the diameter quite than the radius, you have the right to likewise deal straight via the diameter from the start and conserve some job-related later. This is based on the reality that $r=fracd2$ (radius is constantly half the diameter).

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

Now you have the right to take the derivative with respect to time.

$$fracdSAdt=2 pi d fracdddt$$

Now just plug in the provided indevelopment and also fix for $fracdddt$.

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

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

Here"s a great webwebsite that describes this problem in a little bit even more information. The numbers are different, however the procedure is similar.


edited Jan 30 "19 at 6:35
answered Jan 8 "19 at 15:02

Jake OJake O
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