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Well, in physics we got $W = V\cdot I\cdot t \stackrel{R=^V/_I}{\Longrightarrow} \begin{cases} W = I^2\cdot R \cdot t \\ W = V^2\cdot \frac{1}{R}\cdot t \end{cases}$

In the first case the slope of $f(R)=W$ is a line while in the second case the slope is a hyperbola.

In a similar occasion let $f(x) = y = ax$ where $x = \dfrac{a}{b}$. Then by multiplying the numerator and the denominator with $b$ we got $f(x) = x^2b$. In this case the slope of $f(x) = y$ is a parabola while in the former case it was a line. Why is that?

user302454
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1 Answers1

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Notice, since $I=\frac{V}{R}$ is the function of $R$ hence, $$\frac{dI}{dR}=-\frac{V}{R^2}$$

In first case, the slope of $W=f(R)=I^2\cdot R\cdot t$ is given by differentiating $W$ w.r.t. $R$ using product rule $$\frac{dW}{dR}=Rt\frac{dI^2}{dR}+I^2t\frac{dR}{dR}$$$$=2IRt\frac{dI}{dR}+I^2t$$ $$=2\left(\frac{V}{R}\right) Rt\left(-\frac{V}{R^2}\right)+\left(\frac{V}{R}\right)^2t$$ $$=\color{red}{-\frac{V^2t}{R^2}}$$

In the second case, the slope of $W=f(R)=V^2\cdot \frac{1}{R}\cdot t$ is given by differentiating $W$ w.r.t. $R$ as follows $$\frac{dW}{dR}=V^2t\frac{d}{dR}\left(\frac{1}{R}\right)=V^2t\left(-\frac{1}{R^2}\right)=\color{red}{-\frac{V^2t}{R^2}}$$

Thus, in both the cases, the results are same i.e. in both the cases, the slope is a hyperbola.

Harish Chandra Rajpoot
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