Differential Calculus-2
Differential Calculus-2
DAILY_CHALLENGE
Grade 12
Question:
$l_1$ and $l_2$ are lengths of side of two variable squares $S_1$ and $S_2$ respectively for $l_i = l_1^2 + l_2^2 + 6$ at $l_2 = 1$. If rate of change of area of $S_2$ with respect to area of $S_1$ is equal to $\frac{1}{8m}$, then $m = $.
Step-by-Step Solution
Key Concept: Differentiate a composite relation involving multiple variables and use chain rule to find the ratio of rates.
Given $\sqrt{A_1} = \sqrt{A_2} + 4y^2 + 6$, we differentiate with respect to $t$: $\frac{1}{2\sqrt{A_1}} \cdot \frac{dA_1}{dt} = \frac{1}{2\sqrt{A_2}} \cdot \frac{dA_2}{dt} + 8y \frac{dy}{dt}$. Substituting known values yields $\frac{dA_2}{dt} / \frac{dA_1}{dt} = 1/32$.
Correct Answer: 4