- #1
crookesm
- 5
- 0
Imagine a cube that 'grows' by five cubic inches per week. How fast is its surface area increasing when the length of one of its sides is seven inches?
I know that the derivative of volume (V) with respect to time (t) is 5, e.g:
[tex] \frac{dV}{dt} = 5[/tex]
To calculate the surface area of a cube from a given volume I would use:
[tex] S=6(\sqrt[3]{V})^2 [/tex]
Therefore, would I need to calculate the derivate of S with respect to V in order to reach my goal of deriving S with respect to t?
Or am I going totally in the wrong direction?
I know that the derivative of volume (V) with respect to time (t) is 5, e.g:
[tex] \frac{dV}{dt} = 5[/tex]
To calculate the surface area of a cube from a given volume I would use:
[tex] S=6(\sqrt[3]{V})^2 [/tex]
Therefore, would I need to calculate the derivate of S with respect to V in order to reach my goal of deriving S with respect to t?
Or am I going totally in the wrong direction?