Question 4

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer

Let *x* be the length of a side and *V* be the volume of the cube. Then,

*V* = *x*^{3.}

\begin{align}\therefore \frac{dV}{dt}=3x^2.\frac{dx}{dt}\;\;\;[By\; Chain \;Rule]\end{align}

It is given that,

\begin{align} \frac{dx}{dt}=3 \;cm^2/s\end{align}

\begin{align}\therefore \frac{dV}{dt}=3x^2.(3) = 9x^2\end{align}

Thus, when *x* = 10 cm,

\begin{align} \frac{dV}{dt}=9 (10)^2=900 \;cm^3/s\end{align}

Hence, the volume of the cube is increasing at the rate of 900 cm^{3}/s when the edge is 10 cm long.

harsh patel
2018-04-15 11:25:30

why dv/dt=3x^2.dx/dt

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