Q109 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
A point P is moving on the curve $x^3 y^4=2^7$. The $x$-coordinate of $P$ is decreasing at the rate of 8 units per second. When the point $P$ is at $(2,2)$, the $y$-coordinate of $P$
- Correct answerAincreases at the rate of 6 units per second
- Bdecreases at the rate of 6 units per second
- Cincreases at the rate of 4 units per second
- Ddecreases at the rate of 4 units per second
Solution
$x^3 y^4=2^7$, Differentiate w.r.t. $x$
$\begin{aligned}
& 3 x^2 y^4+4 x^3 y^3 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-3 y}{4 x} \Rightarrow d y=\frac{-3 y}{4 x} d x \\
& \Rightarrow d y=\frac{6 y}{x} \Rightarrow(d y)_{(2,2)}=6
\end{aligned}$
Hence $y$-coordinate increases at the rate of 6 units $/ \mathrm{sec}$.
Hence $y$-coordinate increases at the rate of 6 units $/ \mathrm{sec}$.
Hence, the correct answer is option: 'A'.