MHT CET · Maths · Application of Derivatives
The slope of the normal to the curve \(x=\sqrt{t}\) and \(y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}\) at \(\mathrm{t}=4\) is
- A \(\frac{-17}{4}\)
- B \(\frac{4}{17}\)
- C \(\frac{-4}{17}\)
- D \(\frac{17}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{-4}{17}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{1+\frac{1}{2 \mathrm{t}^{\frac{3}{2}}}}{\frac{1}{2 \sqrt{\mathrm{t}}}}=\frac{2 \mathrm{t}^{\frac{3}{2}}+1}{\mathrm{t}} \)
\( \therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=4}= \frac{2(4)^{\frac{3}{2}}+1}{4}=\frac{16+1}{4}=\frac{17}{4}\)
\(\therefore\) Slope of normal at \(\mathrm{t}=4\) is \(-\frac{1}{\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{t=4}}=-\frac{4}{17}\)
\( \therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{\mathrm{t}=4}= \frac{2(4)^{\frac{3}{2}}+1}{4}=\frac{16+1}{4}=\frac{17}{4}\)
\(\therefore\) Slope of normal at \(\mathrm{t}=4\) is \(-\frac{1}{\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{t=4}}=-\frac{4}{17}\)
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