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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\text{t}+\frac{1}{\text{t}},\text{ y}=\text{t}-\frac{1}{\text{t}}$
✓

Answer

Consider, $\text{x}=\text{t}+\frac{1}{\text{t}}$ and $\text{y}=\text{t}\frac{1}{\text{t}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1+\Big(-\frac{1}{\text{t}^2}\Big)$ and $\frac{\text{dy}}{\text{dt}}=1-\Big(-\frac{1}{\text{t}^2}\Big)$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=1-\frac{1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=1+\frac{1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{t}^2-1}{\text{t}^2}$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{t}^2+1}{\text{t}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\text{t}^2+\frac{1}{\text{t}^2}}{\text{t}^2-\frac{1}{\text{t}^2}}$
$=\frac{\text{t}^2+1}{\text{t}^2-1}$

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