CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The volume of a cube is increasing at the rate of $$8 cm^3/s$$. How fast is the surface area increasing when the length of an edge is $$12 cm$$?


Solution

Let $$x$$ be the length of a side, $$V$$ be the volume, and $$S$$ be the surface area of the cube.

Then, $$V =x^3$$  and $$S = 6x^2$$ 
It is given that $$\cfrac {dV}{dt}=8  cm^3/s$$.
Then, by using the chain rule, we have: 
$$\therefore\displaystyle   8=\frac {dV}{dt}=\frac {d}{dt}(x^3)\cdot \frac{d}{dx}=3x^2\cdot \frac {dx}{dt}$$

$$\Rightarrow\displaystyle  \frac {dx}{dt}=\frac {8}{3x^2}.........(1)$$
Now, $$\displaystyle  \frac {dS}{dt}=\frac {d}{dt}(6x^2)\cdot \frac{d}{dx}=(12x)\cdot \frac {dx}{dt}$$ [By chain rule]

$$\displaystyle =12x\cdot \frac {dx}{dt}=12x\cdot \left ( \frac {8}{3x^2} \right )=\frac {32}{x}$$

Thus, when $$x = 12$$ cm,  $$\displaystyle  \frac {dS}{dt}=\frac {32}{12} cm^2/s=\frac {8}{3} cm^2/s.$$
Hence, if the length of the edge of the cube is $$12$$ cm, then the surface area is increasing  at the rate of $$\cfrac {8}{3} cm^2/s$$.

Mathematics
RS Agarwal
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image