the rate of an expanding rectangle is increasing at a rate of 48 cmsq/sec. The length of the rectangle is always equal to the square of the breadth. At what rate is the length increasing at the instant when the breadth is 4.5 cm .
Asked
Active
Viewed 2,099 times
2 Answers
2
HINT;
If the breadth $=a$ cm, length$ =a^2$ cm
So, the area is $=a^3$ cm sq
Now, $\frac{d(a^3)}{dt}=48$ cm sq $\implies 3a^2\cdot \frac{da}{dt}=48$ cm sq
We need $\frac{d(a^2)}{dt}_{(\text{at } a=4.5 cm)}=2a\cdot\frac{d a}{dt}_{(\text{at } a=4.5 cm)}$

lab bhattacharjee
- 274,582
0
Let the area at time $t$ be $A=A(t)$, and let the breadth at time $t$ be $x=x(t)$. Then the length at time $t$ is $x^2$, and therefore $$A=x^3.$$ Differentiate with respect to $t$, using the Chain Rule. We get $$\frac{dA}{dt}=3x^2\frac{dx}{dt}.\tag{1}$$ Now freeze the situation at the instant that $x=4.5$. We know $\frac{dA}{dt}$, and we know $x$, so from (1) we can find $\frac{dx}{dt}$ at that time.

André Nicolas
- 507,029