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Question Video: Finding the Rate of Change of a Resistance given the Rates of Change of the Voltage and the Current across It Using Related Rates Mathematics • Higher Education

In a closed electric circuit, ๐‘‰ is the potential difference measured in volts, ๐ผ is the current intensity measured in amperes, and ๐‘… is the resistance measured in ohms. If the potential difference increases at a rate of 6 volts per second, and the current intensity decreases at a rate of 1/2 A/S, find the rate of change of the resistance when ๐‘‰ = 20 volts and ๐ผ = 5 amperes.

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Video Transcript

In a closed electric circuit, ๐‘‰ is the potential difference measured in volts, ๐ผ 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 ๐ผ equals five amperes.

Thereโ€™s a lot of information here. So letโ€™s just begin by writing expressions for what we already know. First, weโ€™re told that the potential difference, thatโ€™s ๐‘‰, increases at a rate of six volts per second. In other words, weโ€™re 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 d๐‘ก. Now, since this increases at a rate of six volts per second, d๐‘‰ by d๐‘ก must be positive six.

Similarly, weโ€™re also given information about the rate of change of current intensity ๐ผ. This decreases at a rate of one-half amperes per second, so d๐ผ by d๐‘ก must be negative one-half. Itโ€™s negative since the currentโ€™s intensity is decreasing over time.

The question wants us to find the rate of change of the resistance when ๐‘‰ is 20 and ๐ผ is five. Since ๐‘… is the resistance, in other words, the question wants us to find d๐‘… by d๐‘ก.

So we need to find a way to link d๐‘‰ by d๐‘ก, d๐ผ by d๐‘ก, and d๐‘… by d๐‘ก. Well, to do so, weโ€™re going to use the following formula: ๐‘‰ equals ๐ผ times ๐‘…. Since weโ€™re interested in finding the rate of change of the resistance, letโ€™s rewrite this as ๐‘… equals ๐‘‰ over ๐ผ. Then it follows that we can find an expression for d๐‘… by d๐‘ก in terms of ๐‘‰ and ๐ผ by using the quotient rule for differentiation. In other words, d๐‘… by d๐‘ก is ๐ผ times d๐‘‰ by d๐‘ก minus ๐‘‰ times d๐ผ by d๐‘ก over ๐ผ squared.

Now weโ€™ve already formed numerical expressions for d๐‘‰ by d๐‘ก and d๐ผ by d๐‘ก. But weโ€™re also told that ๐‘‰ equals 20 and ๐ผ equals five at the exact instance that weโ€™re interested in. So weโ€™re able to find d๐‘… by d๐‘ก at this exact instance by substituting all of these values in. ๐ผ times d๐‘‰ by d๐‘ก then is five times six. Then we subtract ๐‘‰ times d๐ผ by d๐‘ก. Thatโ€™s 20 times negative one-half. And of course, we divide that by ๐ผ squared, which is five squared.

Evaluating each expression, and we get 30 plus 10 over 25, which gives us d๐‘… by d๐‘ก 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 d๐‘ก 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 ๐ผ is five is eight-fifths ohms per second.

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