Rate of change question

Dec 2008
509
2
Hi

I need help on the following question:
1) A machine component has the form of a solid right circular cylinder. If the cylinder is uniformly heated both the radius 'r', and length 'L', increase which increases the surface area and volume of the cylinder.

If the radius is increasing at a rate of 0.60559 cm/hour and length increases at a rate of 0.13985 cm/hour at a time when the radius is 2 cm and length is 17cm , then the rate at which the surface area S=2pr^2+2prL is changing with time is approximately:


\(\displaystyle \frac{dr}{dt} = 0.60559 cm/hr\)

\(\displaystyle \frac{dL}{dt} = 0.13958 cm/hr
\)
r = 2cm

L = 17 cm

\(\displaystyle \frac{dS}{dt} = 2\pi(2r\frac{dr}{dt}) + 2\pi(r\frac{dL}{dt} + L\frac{dr}{dt})\)

\(\displaystyle \frac{dS}{dt} = 2\pi(2 * 2 * 0.60559) + 2\pi(2 * 0.13985 + 17 * 0.60559)\)

\(\displaystyle \frac{dS}{dt} = 81.663
\)
I don't know why this is incorrect?

The correct answer is 12.9971

P.S
 
Jan 2010
354
173
Your answer appears to be correct to me.
 
Dec 2009
872
381
1111
Hi

I need help on the following question:
1) A machine component has the form of a solid right circular cylinder. If the cylinder is uniformly heated both the radius 'r', and length 'L', increase which increases the surface area and volume of the cylinder.

If the radius is increasing at a rate of 0.60559 cm/hour and length increases at a rate of 0.13985 cm/hour at a time when the radius is 2 cm and length is 17cm , then the rate at which the surface area S=2pr^2+2prL is changing with time is approximately:


\(\displaystyle \frac{dr}{dt} = 0.60559 cm/hr\)

\(\displaystyle \frac{dL}{dt} = 0.13958 cm/hr
\)
r = 2cm

L = 17 cm

\(\displaystyle \frac{dS}{dt} = 2\pi(2r\frac{dr}{dt}) + 2\pi(r\frac{dL}{dt} + L\frac{dr}{dt})\)

\(\displaystyle \frac{dS}{dt} = 2\pi(2 * 2 * 0.60559) + 2\pi(2 * 0.13985 + 17 * 0.60559)\)

\(\displaystyle \frac{dS}{dt} = 81.663
\)
I don't know why this is incorrect?

The correct answer is 12.9971

P.S
Dear Paymemoney,

I think your answer is correct. I checked and rechecked but coud'nt find anything wrong.