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?(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Rates of change - confused (again!)

Loading...

Similar Threads for Rates change confused | Date |
---|---|

I Rate of change of area under curve f(x) = f(x) | Jan 2, 2018 |

I Gradient Vector- largest possible rate of change? | Mar 8, 2016 |

Rate of change of area of circle in respect to radius | Feb 25, 2014 |

Rate of change of area of a square with respect to its perimeter | May 27, 2013 |

**Physics Forums - The Fusion of Science and Community**