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

Hint: 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.