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.

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$$S=\pi x^2 =-\frac32\pi 8 My answer is wrong. Have the right to some one help me understand? calculus share cite follow edited Mar 16 "15 in ~ 11:51 N. F. Taussig 60.8k1212 yellow badges4848 silver- badges6666 bronze title request Mar 16 "15 in ~ 11:14 CetshwayoCetshwayo 2,8821010 yellow badges4242 silver badges7474 bronze title \endgroup include a comment | ## 2 answers 2 active earliest Votes 3 \begingroup We have that the surface ar area formula is:$$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$$Therefore ours diameter decreases at a price of \frac632\pi re-publishing cite follow answer Mar 16 "15 at 11:22 Varun IyerVarun Iyer 5,8581313 silver badges2929 bronze title \endgroup add a comment | 0 \begingroup Since the question is asking about the diameter quite than the radius, girlfriend can additionally deal directly with the diameter indigenous the beginning and also save some job-related later. This is based on the truth that r=\fracd2 (radius is always half the diameter).$$SA=4 \pi r^2SA=4 \pi \bigg( \fracd2 \bigg)^2SA=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. https://jakesrwandachamber.orglessons.com/derivatives/related-rates-sphere-surface-area-problem/ re-publishing point out monitor edited january 30 "19 at 6:35 answered jan 8 "19 in ~ 15:02 Jake OJake O 17655 bronze badges$\endgroup\$
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