TS EAMCET · Maths · Application of Derivatives
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 \(P\)
- A increases at the rate of 6 units per second
- B decreases at the rate of 6 units per second
- C increases at the rate of 4 units per second
- D decreases at the rate of 4 units per second
Answer & Solution
Correct Answer
(A) increases at the rate of 6 units per second
Step-by-step Solution
Detailed explanation
\(x^3 y^4=2^7\), 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…
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