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An edge of a variable cube is increasing at the rate of 3 cm/s. How fa...

An edge of a variable cube is increasing | Class 12 Mathematics Chapter Application of Derivatives, Application of Derivatives NCERT Solutions

Question:

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 cm³/s when the edge is 10 cm long.

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Comments

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

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

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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , 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 i....

An edge of a variable cube is increasing | Class 12 Mathematics Chapter Application of Derivatives, Application of Derivatives NCERT Solutions

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