Question

Determine how fast the length of an edge of a cube is changing at the moment when length of the edge is 5cm and the volume of the cube is decreasing at a rate of $100 c{{m}^{3}}/\sec$.

Answer 1

In this problem, we have to find the rate of change of edge of a cube whose length of the edge is 5cm. We are given the volume of the cube is $100c{{m}^{3}}/\sec$. We can now assume the edge as x, then its rate of change will be $\dfrac{dx}{dt}$. We can see that we are given $c{{m}^{3}}$ which indicates the rate of change in volume per second. We have to find $\dfrac{dx}{dt}$ by differentiating the given values to find the answer.

Complete step by step solution:
We know that the given edge of the cube is 5cm.
We have to find the rate of change of the edge $\dfrac{dx}{dt}$.
We know that the given rate of change of volume is,
$\dfrac{dV}{dt}=100c{{m}^{3}}/\sec$ …….. (1)
Here volume of the cube is
$\Rightarrow V={{x}^{3}}$
We can now differentiate the volume, V with respect to time, t, we get
$\Rightarrow \dfrac{dV}{dt}=3{{x}^{2}}\dfrac{dx}{dt}$
We can now simplify the above step, we get
$\Rightarrow \dfrac{dx}{dt}=\dfrac{1}{3{{x}^{2}}}\dfrac{dV}{dt}$
We can now substitute the given edge value and the (1) in the above step, we get
$\Rightarrow \dfrac{dx}{dt}=100\left( \dfrac{1}{3{{\left( 5 \right)}^{2}}} \right)=\dfrac{4}{3}cm/\sec$
We are given that the edges are decreasing, so the answer will be negative, we get
$\Rightarrow \dfrac{dx}{dt}=100\left( \dfrac{1}{3{{\left( 5 \right)}^{2}}} \right)=-\dfrac{4}{3}cm/\sec$

Therefore, the length of the edge of the cube will be decreasing at a speed of $-\dfrac{4}{3}cm/\sec$

Note: We should know that $\dfrac{dV}{dt}$ is the rate of change of volume with respect to time and $\dfrac{dr}{dt}$ is the rate of change of radius with respect to time. Here the given basketball is nothing but a sphere whose volume is $\Rightarrow V=\dfrac{4}{3}\pi {{r}^{3}}$. We should also mention the unit in the answer part.