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  • Find dy/dx of each of the functions expressed in parametric form x =e^theta

Find \frac{dy}{dx} of each of the functions expressed in parametric form

\mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right), \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)

Answers (1)

We have two equations \mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right), \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)

Now, differentiate w.r.t θ

So,

\\ \mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right)\right] \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right)\left(\mathrm{e}^{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}+\theta+\frac{1}{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)...(i)

Also,

\\ \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\ \frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)\right] \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{-\theta}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)-\left(\theta-\frac{1}{\theta}\right)\left(\mathrm{e}^{-\theta}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)_{-\mathrm{ }(\mathrm{ii})}

\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)}{\mathrm{e}^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)

Posted by

infoexpert22

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