Calculus help?

Question

If a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decrases when the diameter is 14 cm.

im having trouble with the chain rule

i got -175.9291886 but once again i was wrong

Answer 1

You don't need the chain rule on this one.

The surface area of the snowball (spherical approximation) is:

SA = 4 pi r^2 = 4 pi (D/2)^2 = pi D^2, where D is the diameter.

Therefore d SA/ d D = 2 pi D

or, d SA = 2 pi D d D

so d SA/dt = 2 pi D d D/dt

so d D/dt = d SA/dt divided by 2 pi D

You are given that D = 14 cm and that d SA /dt = 1cm^2 / min

So when the diameter is 14 cm we have:

d D/dt = 1 cm^2 /min divided by 2*pi*14 cm

= 0.01137 cm / min

If you are having trouble with the chain rule, I suggest you use the following link which explains it quite clearly with examples.

http://en.wikipedia.org/wiki/Chain_rule

Answer 2

i might say shape is area of engineering meaning that a sturdy carry close on math is needed, pre-cal is is like algebra to the subsequent point with the Pi chart and understanding approximately radians and perspective measures (which seems substantial) yet Calculus is the study of derivatives, (the slope of a line at a undeniable factor) style of pointless no remember what container of workd your going to, in case you inquire from me.

Answer 3

surface area=4pir^2=4pi(d^2)/4
dA/dt=2piddd/dt
1=2pi*14*dd/dt
dd/dt=(1/28pi)cm/min
the rate at which the diameter
is(1/28pi)cm/min

Answer 4

SA= 4(Pi)r^2 d(SA)/dt= 8(Pi)r(dr/dt)

d(SA)/dt= -1

r=D/2=14/2=7

dr/dt= (1/2)dD/dt

-1=8(Pi)(7)(dr/dt)

dr/dt= -1/(8*7*Pi)

dD/dt= 2dr/dt= -1/(4*7*Pi)