Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 2 • Question 25

A conic reservoir is filled with water. The graph shows the height of water in the reservoir for 11 minutes. Use the graph to work out an estimate for the rate at which the height is increasing at 3 minutes. You must show your working.

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Video Transcript

A conic reservoir is filled with water. The graph shows the height of water in the reservoir for 11 minutes. Use the graph to work out an estimate for the rate at which the height is increasing at three minutes. You must show your working.

So reading the question, we can see that we want to work out an estimate for the rate at which the height is increasing at three minutes. So I’ve marked on this point on our graph. So what we’re looking for at this point is in fact a rate of change. And the rate of change of a graph is gonna be the gradient, because the rate of change is the rate at which the height is increasing over a set amount of time. So therefore, it’s the change in 𝑦 over the change in 𝑥, which is the gradient.

However, we don’t have the tools to work out the rate of change of the gradient at a particular point. If we did, we’d be using something like calculus, and that would be moving on into A level. So what do we do to find the gradient at this point?

Well, what we do to work out a gradient at this point is we draw a tangent. And this tangent to the curve is gonna help us to estimate the gradient or the rate of change at the point that we’re looking for. So what we’re going to do is work out the gradient of our tangent.

It is important to remember, however, that when you draw your tangent, so your straight line, it must follow the curve of the graph that you’re looking at. So what I’ve drawn is some bad examples of lines that could’ve been drawn. So the pink line is far too steep. The blue line is far too shallow. And neither of these give a fair representation of what the gradient would be at that point as they do not follow the graph.

Well, next, to find the gradient of the line, I’m gonna choose two points on the line. So the two points on the tangent that I’ve chosen they can be anywhere, because the gradient of the tangent or the gradient of the straight line will not change. It will remain constant. But what I’ve done is I’ve made sure that it’s easy to read off both the 𝑥- and 𝑦-values, so in this case the time and the height, of the two points I’ve chosen.

So I’ve got one at zero, three and one at six, seven. And as I’ve already mentioned, to find the gradient, it’s gonna be the change in 𝑦 divided by the change in 𝑥. And remembering that we’re trying to find the gradient, because the gradient is the rate of change, which is what we’re looking for in this question.

Well, as we can see from the graph, our change in 𝑦 is going to be four. And that’s because we have seven and three. And seven take away three is four. And the change in 𝑥 is going to be six. And that’s cause we go from zero to six. And six take away zero is just six.

Okay, great, so now we’ve got our change in 𝑦 and our change in 𝑥. So what we can do is find our gradient. So therefore, we can say that the gradient is gonna be four over six or four-sixths, which can be simplified by dividing both the numerator and denominator by two. So when we do that, we get two-thirds. And that’s cause four divided by two is two and six divided by two is three. And we chose two because it was the highest common factor of four and six.

So remembering once again that the rate of change is equal to the gradient, we can say therefore that the rate of height increase is gonna be equal to two-thirds of a centimeter per minute. And we can remember this is the rate of change at three minutes. And we found that by first drawing a tangent to the curve at that point. And that’s to estimate the gradient. Then we found the gradient using change in 𝑦 divided by change in 𝑥. And then it gives us the answer that we got.

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