-2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Welcome to Math Stack Exchange. You are supposed to show what you have tried and specify your difficulties. $\endgroup$ – Danilo Gregorin Sep 25 '17 at 17:52
1
$\begingroup$

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.
$\endgroup$
0
$\begingroup$

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 $$ R'(t) = \frac{V'(t) - I'(t)R(t)}{I(t)} =\frac{3 + \frac{1}{4}\cdot \frac{5}{9}}{5} = \frac{3\cdot 36 + 5}{5\cdot 36} = \frac{113}{180}.$$

$\endgroup$
0
$\begingroup$

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 $R'(t) = \color{red}{\frac{69}{100}} \Omega/s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.