Math Help - Rate of change question

1. Rate of change question

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:

$\frac{dr}{dt} = 0.60559 cm/hr$

$\frac{dL}{dt} = 0.13958 cm/hr
$

r = 2cm

L = 17 cm

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

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

$\frac{dS}{dt} = 81.663
$

I don't know why this is incorrect?

P.S

3. Originally Posted by Paymemoney
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:

$\frac{dr}{dt} = 0.60559 cm/hr$

$\frac{dL}{dt} = 0.13958 cm/hr
$

r = 2cm

L = 17 cm

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

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

$\frac{dS}{dt} = 81.663
$

I don't know why this is incorrect?