The voltage $V$ (volts), current $I$ (amperes), and resistance $R$ (ohms) of an electric circuit are related by the equation $V = IR$.

Suppose that $V$ is increasing at a rate of $3$ volt/sec while $I$ is decreasing at a rate of $-\frac{1}{4}$ amp/sec. Let $t$ denote time in seconds.

Determine the rate at which $R$ is changing when $V$ = 9 volts and $I$ = 5 amperes.

• Welcome to Math Stack Exchange. You are supposed to show what you have tried and specify your difficulties. – Danilo Gregorin Sep 25 '17 at 17:52

HINT

So you have $$V(t) = I(t) R(t).$$

1. Compute $V'(t)$ using product rule.
2. You are given $V'(t), I'(t)$ and $V,I$ are specified as well.
3. Plug into your result in (1) to compute what you need.

We are given that $$V(t) = I(t) R(t).$$ Differentiating with respect to time, we get (via the product rule) $$\frac{\mathrm{d}V}{\mathrm{d}t}(t) = \frac{\mathrm{d}I}{\mathrm{d}t}(t) \cdot R(t) + I(t)\cdot \frac{\mathrm{d}R}{\mathrm{d}t}(t).$$ Alternatively, in a more Newton-esque notation, this is $$V'(t) = I'(t)R(t) + I(t)R'(t).$$ Our goal is to find the rate at which $R$ is changing, i.e. to find $R'$. Solving the above, we obtain $$R'(t) = \frac{V'(t) - I'(t)R(t)}{I(t)}. \tag{\ast}$$ All of the data on the right-hand side are given in the statement of the problem---it just requires a little interpretation.

We are told that voltage is increasing at a rate of 3 volts per second. But this corresponds to an instantaneous rate of change, i.e. the derivative. Therefore (suppressing units---this is not generally a good idea, but I am lazy) $$V'(t) = 3.$$ By similar reasoning, we have $$I'(t) = -\frac{1}{4}.$$

Finally, we are also told that $$V(t) = 9 \qquad\text{and that}\qquad I(t) = 5.$$ From the first equation, this implies that $$R(t) = \frac{I(t)}{V(t)} = \frac{5}{9}.$$ Substituting all of these back into ($\ast$), we obtain $$R'(t) = \frac{V'(t) - I'(t)R(t)}{I(t)} =\frac{3 + \frac{1}{4}\cdot \frac{5}{9}}{5} = \frac{3\cdot 36 + 5}{5\cdot 36} = \frac{113}{180}.$$

Hint: $$R(t)=\frac{V(t)}{I(t)} \implies R'(t) = \frac{V'(t)I(t)-V(t)I'(t)}{I(t)^2}.$$ You should get $R'(t) = \color{red}{\frac{69}{100}} \Omega/s$.