Question Video: Finding the Rate of Change in the Area of an Expanding Rectangle Using Related Rates

The length of a rectangle is increasing at a rate of 15 cm/s and its width at a rate of 13 cm/s. Determine the rate at which the area of the rectangle increases when the length of the rectangle is 25 cm and its width is 12 cm.

02:50

Video Transcript

The length of a rectangle is increasing at a rate of 15 centimetres per second and its width at a rate of 13 centimetres per second. Determine the rate at which the area of the rectangle increases when the length of the rectangle is 25 centimetres and its width is 12 centimetres.

Remember, we should always try to identify what information we’ve been given and what we’re being asked to find. We know that the length of the rectangle is increasing at a rate of 15 centimetres per second. This tells us two things. Firstly, length is a function of time. It also tells us though that the rate of change, which is simply the derivative with respect to time, is 15. So, we can say that d𝑙, where 𝑙 is the length, by d𝑡 is equal to 15.

Similarly, we’re told the width of the rectangle is increasing at a rate of 13 centimetres per second. So, width is also a function of time. And since the rate of change is the derivative with respect to time, d𝑤, where 𝑤 is the width, by d𝑡 is equal to 13.

Now we’re looking to find the rate at which the area is increasing. So, we use two pieces of information. Firstly, we know that the area of a rectangle is given by its width multiplied by its length. But we’re trying to find the rate of change of the area, that’s d𝐴 by d𝑡. Let’s say area is 𝑤𝑙, width times length, then we can say that d𝐴 by d𝑡 is the derivative of 𝑤 times 𝑙 with respect to time.

We’re trying to find the derivative of a product of two functions of 𝑡. So, we can use the product rule to evaluate the rate of change of the area of our rectangle. This says that if 𝑢 and 𝑣 are differentiable functions, then the derivative of 𝑢 times 𝑣 with respect to 𝑥 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. And what does this mean for the derivative of the area with respect to time?

Well, we can say it’s 𝑤 times d𝑙 by d𝑡 plus 𝑙 times d𝑤 by d𝑡. And actually, we have all of these things. We want to find the rate at which the area of the rectangle increases when its width is 12. So, 𝑤 times d𝑙 by d𝑡 is 12 times 15. And we want to find the rate of change of the area when the length is 25. So, 𝑙 times d𝑤 by d𝑡 is 25 times 13. 12 multiplied by 15 plus 25 multiplied by 13 is 505.

Since the width and length of the rectangle are measured in centimetres, and the rate of change of these measurements is centimetres per second, we know that the rate at which the area of the rectangle increases when the length of the rectangle is 25 centimetres and its width is 12 centimetres is 505 square centimetres per second.

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