Video: Finding the Rate of Change of the Distance between a Fixed Point and a Point Moving on the Curve of a Root Function Using Related Rates

A point is moving on the curve of the function 𝑓(π‘₯) = √(π‘₯Β² + 2). If its π‘₯-coordinate increases at a rate of 9√15 cm/s, find the rate of change of the distance between this point and the point (1, 0) when π‘₯ = 3.

06:20

Video Transcript

A point is moving on the curve of the function 𝑓 of π‘₯ is equal to the square root of π‘₯ squared plus two. If its π‘₯-coordinate increases at a rate of nine square root of 15 centimeters per second, find the rate of change of the distance between this point and the point one, zero when π‘₯ is equal to three.

The question tells us that a point is moving on the curve of the function 𝑓 of π‘₯ is equal to the square root of π‘₯ squared plus two. It also tells us that its π‘₯-coordinate is increasing at a rate of nine square root of 15 centimeters every second. The question wants us to find the rate of change of the distance between this point and the point one, zero when π‘₯ is equal to three. Since the question wants us to find the rate of change of the distance between this point and the point one, zero, let’s calculate a formula for the distance between these two points. Since the point lies on the curve of the function 𝑓 of π‘₯ is equal to the square root of π‘₯ squared plus two, its 𝑦-coordinate is going to be the output of this function. In other words, its Cartesian coordinates are π‘₯, square root of π‘₯ squared plus two.

So we want to find an expression for the distance between the point one, zero and the point π‘₯, square root of π‘₯ squared plus two. We can do this by constructing a right-angled triangle and then using the Pythagorean theorem. Since we’re interested at the rate of change at the point when π‘₯ is equal to three, we see that π‘₯ is greater than one and the square root of π‘₯ squared plus two is greater than zero at this point. So we’ll sketch the point π‘₯, square root of π‘₯ squared plus two up into the right of our point one, zero. We can then subtract the π‘₯-coordinates of these two points to find the base of our right-angled triangle. And we can subtract the 𝑦-coordinates to find the height of this right-angled triangle. This gives us a base of length π‘₯ minus one and a height of length square root of π‘₯ squared plus two.

So by using the Pythagorean theorem, we have our distance squared is equal to π‘₯ minus one squared plus the square root of π‘₯ squared plus two squared. Distributing our exponent over the parentheses gives us π‘₯ squared minus two π‘₯ plus one. And the square root of π‘₯ squared plus two all squared is just π‘₯ squared plus two. We can simplify this further by adding our π‘₯ squared terms and adding our constants together. This gives us two π‘₯ squared minus two π‘₯ plus three. Finally, we’ll take square roots of both sides of this equation. Since our distance is positive, this gives us that 𝐷 is equal to the square root of two π‘₯ squared minus two π‘₯ plus three. And so we found an equation for our distance function as a function of π‘₯.

But the question wants us to find the rate of change of our distance with respect to time when π‘₯ is equal to three. But we have a function of π‘₯, so we’re going to have to apply the chain rule. The chain rule tells us if 𝑦 is some function of 𝑒 and 𝑒 is some function of π‘₯, then we can calculate the derivative of 𝑦 with respect to π‘₯. But first calculating the derivative of 𝑦 with respect to 𝑒 and multiplying this by the derivative of 𝑒 with respect to π‘₯. So by using the chain rule, we can calculate the derivative of our distance function with respect to time. But first calculating the derivative of our distance function with respect to π‘₯ and then multiplying this by the derivative of π‘₯ with respect to time.

The question tells us the π‘₯-coordinate is increasing at a rate of nine root 15 centimeters per second. This is the same as saying the derivative of π‘₯ with respect to time is equal to nine root 15. So we’ve already found dπ‘₯ by d𝑑. This means all we need to do now is find the derivative of our distance function with respect to π‘₯. So we need to calculate the derivative of two π‘₯ squared minus two π‘₯ plus three all raised to the power of a half with respect to π‘₯. We’re now trying to differentiate the composition of two functions, so we can do this again by using the chain rule. If we set 𝑒 equal to two π‘₯ squared minus two π‘₯ plus three, which is our inner function, then we have that 𝐷 is equal to 𝑒 to the power of a half. But 𝑒 is a function of π‘₯.

So by using the chain rule, we can calculate the derivative of our distance function with respect to π‘₯ by first calculating the derivative of our distance function with respect to 𝑒 and then multiplying this by the derivative of 𝑒 with respect to π‘₯. To calculate the derivative of our distance function with respect to 𝑒, we just need to calculate the derivative of 𝑒 to the power of a half with respect to 𝑒. We can do this by using the power rule for derivatives. We multiply by the exponent of a half and then reduce the exponent by one. This gives us a half 𝑒 to the power of negative a half.

Now, to calculate d𝑒 by dπ‘₯, we need to differentiate 𝑒 is equal to two π‘₯ squared minus two π‘₯ plus three with respect to π‘₯. We’ll again use the power rule for differentiation. The derivative of two π‘₯ squared is four π‘₯, and we subtract the derivative of two π‘₯, which is two. And finally, the derivative of the constant three is just equal to zero. Finally, we’ll substitute 𝑒 is equal to two π‘₯ squared minus two π‘₯ plus three. And then bringing our negative exponent down into the denominator gives us the derivative of our distance function with respect to π‘₯ is equal to four π‘₯ minus two divided by two multiplied by the square root of two π‘₯ squared minus two π‘₯ plus three.

We are now ready to use the chain rule to find an expression for the derivative of our distance function with respect to time. We’ve calculated the derivative of our distance function with respect to π‘₯ and we’ve calculated the derivative of π‘₯ with respect to 𝑑. This gives us the derivative of our distance function with respect to time is equal to four π‘₯ minus two multiplied by nine root 15 all divided by two multiplied by the square root of two π‘₯ squared minus two π‘₯ plus three.

Finally, the question wants us to find the rate of change of our distance function with respect to time when π‘₯ is equal to three. So we’ll substitute π‘₯ is equal to three into this expression. This gives us four times three minus two multiplied by nine root 15 all divided by two times the square root of two times three squared minus two times three plus three. We have that four times three minus two is 10, and we multiply this by nine root 15 to get a numerator of 90 square root of 15. Similarly, we can evaluate our denominator to get two multiplied by the square root of 15. So we have 90 root 15 divided by two root 15, which we can calculate to give us 45. And since the question tells us that these units are in centimeters per second, we can add units to this answer.

Therefore, we’ve shown that the rate of change of the distance between our point and the point one, zero when π‘₯ is equal to three is 45 centimeters per second.

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