Related Rates

**rhcp1231** · March 10th 2010, 07:50 PM

As a circular griddle is being heated, its diameter changes at a rate of 0.01 cm/min. When the diameter is 30 cm, at what rate is the area of the griddle changing?

I know my formula is V=PI*r^2
and i know since i am given a diameter i would use (h/2) for my radius.

but once i take my derivative of the area formula and plug in my diameter and my DD/DR as .01 cm/min, i am given .942...
the answer is .471

what am i doing wrong?

**Jhevon** · March 10th 2010, 08:10 PM

Originally Posted by rhcp1231

As a circular griddle is being heated, its diameter changes at a rate of 0.01 cm/min. When the diameter is 30 cm, at what rate is the area of the griddle changing?

I know my formula is V=PI*r^2
and i know since i am given a diameter i would use (h/2) for my radius.

but once i take my derivative of the area formula and plug in my diameter and my DD/DR as .01 cm/min, i am given .942...
the answer is .471

what am i doing wrong?

Your formula is: $A = \pi r^2 = \frac {\pi D^2}4$

And so, by differentiating implicitly with respect to time, you get: $\frac {dA}{dt} = \frac {\pi D}2 \cdot \frac {dD}{dt}$

Now, you want $\frac {dA}{dt}$ when $D = 30$ and $\frac {dD}{dt} = 0.01$ , just plug those in and compute.

**rhcp1231** · March 10th 2010, 08:13 PM

Originally Posted by Jhevon

Your formula is: $A = \pi r^2 = \frac {\pi D^2}4$

And so, by differentiating implicitly with respect to time, you get: $\frac {dA}{dt} = \frac {\pi D}2 \cdot \frac {dD}{dt}$

Now, you want $\frac {dA}{dt}$ when $D = 30$ and $\frac {dD}{dt} = 0.01$ , just plug those in and compute.

it was so obvious lol thanks man

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