### Video Transcript

In a closed electric circuit, ๐ is the potential difference measured in volts, ๐ผ is the current intensity measured in amperes, and ๐
is the resistance measured in ohms. If the potential difference increases at a rate of six volts per second, and the current intensity decreases at a rate of one-half amperes per second, find the rate of change of the resistance when ๐ equals 20 volts and ๐ผ equals five amperes.

Thereโs a lot of information here. So letโs just begin by writing expressions for what we already know. First, weโre told that the potential difference, thatโs ๐, increases at a rate of six volts per second. In other words, weโre given information about the rate of change of ๐ with respect to time. When we think about the rate of change of something, we can think about it as its derivative. In other words, if ๐ is the potential difference, then the rate of change of ๐ with respect to time is d๐ by d๐ก. Now, since this increases at a rate of six volts per second, d๐ by d๐ก must be positive six.

Similarly, weโre also given information about the rate of change of current intensity ๐ผ. This decreases at a rate of one-half amperes per second, so d๐ผ by d๐ก must be negative one-half. Itโs negative since the currentโs intensity is decreasing over time.

The question wants us to find the rate of change of the resistance when ๐ is 20 and ๐ผ is five. Since ๐
is the resistance, in other words, the question wants us to find d๐
by d๐ก.

So we need to find a way to link d๐ by d๐ก, d๐ผ by d๐ก, and d๐
by d๐ก. Well, to do so, weโre going to use the following formula: ๐ equals ๐ผ times ๐
. Since weโre interested in finding the rate of change of the resistance, letโs rewrite this as ๐
equals ๐ over ๐ผ. Then it follows that we can find an expression for d๐
by d๐ก in terms of ๐ and ๐ผ by using the quotient rule for differentiation. In other words, d๐
by d๐ก is ๐ผ times d๐ by d๐ก minus ๐ times d๐ผ by d๐ก over ๐ผ squared.

Now weโve already formed numerical expressions for d๐ by d๐ก and d๐ผ by d๐ก. But weโre also told that ๐ equals 20 and ๐ผ equals five at the exact instance that weโre interested in. So weโre able to find d๐
by d๐ก at this exact instance by substituting all of these values in. ๐ผ times d๐ by d๐ก then is five times six. Then we subtract ๐ times d๐ผ by d๐ก. Thatโs 20 times negative one-half. And of course, we divide that by ๐ผ squared, which is five squared.

Evaluating each expression, and we get 30 plus 10 over 25, which gives us d๐
by d๐ก is 40 divided by 25. But of course, we notice we can divide both the numerator and denominator of this fraction by five. So we find d๐
by d๐ก is eight-fifths. Now of course, this is the rate of change of the resistance. And we were told the resistance is measured in ohms. So the rate of change of the resistance when ๐ is 20 and ๐ผ is five is eight-fifths ohms per second.