MHT CET · Maths · Application of Derivatives
An object is moving in the clockwise direction around the unit circle \(x^2+y^2=1\). As it passes through the point \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\), its \(y\)-co-ordinate is decreasing at the rate of 3 units per sec. The rate at which the \(x\)-co-ordinate changes at this point is
- A \(2\) units \(/ \mathrm{sec}\)
- B \(3 \sqrt{3}\) units \(/ \mathrm{sec}\)
- C \(\sqrt 3\) units \(/ \mathrm{sec}\)
- D \(2\sqrt 3\) units \(/ \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(3 \sqrt{3}\) units \(/ \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
Given equation is \(x^2+y^2=1\) Differentiating w.r.t. t, we get
\(\begin{aligned}
& 2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 y \frac{\mathrm{d} y}{\mathrm{dt}}=0 \\
& 2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 y(-3)=0 \\
& \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{6 y}{2 x} \\
& \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{3 y}{x} \\
& \left.\frac{\mathrm{d} x}{\mathrm{dt}}\right|_{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}=\frac{3 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} \\
& =3 \sqrt{3} \text { units/sec }
\end{aligned}\)
\(\begin{aligned}
& 2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 y \frac{\mathrm{d} y}{\mathrm{dt}}=0 \\
& 2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 y(-3)=0 \\
& \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{6 y}{2 x} \\
& \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{3 y}{x} \\
& \left.\frac{\mathrm{d} x}{\mathrm{dt}}\right|_{\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}=\frac{3 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} \\
& =3 \sqrt{3} \text { units/sec }
\end{aligned}\)
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