Video: Modeling with Geometric Sequences

A ball rebounds to π‘Ÿ times its previous height after each bounce. It is observed to rebound to a tenth of its original height on the 6th bounce. What is the value of π‘Ÿ? Round your answer to two decimal places.

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Video Transcript

A ball rebounds to π‘Ÿ times its previous height after each bounce. It is observed to rebound to a tenth of its original height on the sixth bounce. What is the value of π‘Ÿ? Round your answer to two decimal places.

Let’s think about what is actually happening here. The ball is dropped. Each bounce is slightly shorter than the previous bounce. Let’s say the ball originally starts at a height of β„Ž units. Since the ball rebounds to π‘Ÿ times its previous height after each bounce, after the first bounce, it will reach a height of π‘Ÿ times β„Ž units. Then, after the second bounce, it will reach a height of π‘Ÿ times π‘Ÿβ„Ž or π‘Ÿ squared β„Ž. And so, if we take each of the heights of the ball, we see that we’re generating a geometric sequence. This is one where each term is found by multiplying the previous term by a fixed nonzero number. We call this the common ratio. And here we see the common ratio is simply π‘Ÿ.

Now, in fact, there’s a formula that we can use to find any term in a geometric sequence. For a geometric sequence with first term π‘Ž and common ratio π‘Ÿ, the 𝑛th term, π‘Ž sub 𝑛, is π‘Ž times π‘Ÿ to the power of 𝑛 minus one. Now, in this case, our first term is β„Ž, but our common ratio is π‘Ÿ. So the 𝑛th term of our sequence is π‘Ž sub 𝑛 equals β„Ž times π‘Ÿ to the power of 𝑛 minus one.

Now, in fact, we are actually given some information about the height of the ball on the sixth bounce. We know that on the sixth bounce, 𝑛 will be equal to seven since we’re finding the height of the ball after that bounce. And we’re told that it is observed to rebound to a tenth of its original height, so one-tenth of β„Ž. That means the seventh term in our sequence is one-tenth β„Ž. And we can substitute 𝑛 equals seven and π‘Ž sub 𝑛 as one-tenth β„Ž into our formula. We get a tenth β„Ž equals β„Ž times π‘Ÿ to the power of seven minus one or one-tenth β„Ž equals β„Žπ‘Ÿ to the sixth power.

We’re going to divide through by β„Ž. And so we find that one-tenth is equal to π‘Ÿ to the sixth power. We could use logarithms to solve this equation for π‘Ÿ. Alternatively, if our calculators allow it, we simply find the sixth root of both sides. Now, normally, for even powers of π‘Ÿ, we’d be looking to find both the positive and negative root. In this case though, we know that π‘Ÿ has to be a positive number. So π‘Ÿ is equal to the sixth root of one-tenth. That gives us 0.6812 and so on, which is 0.68 correct to two decimal places. And so we say that π‘Ÿ is equal to 0.68.

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