Question Video: A Related Rates Problem on a Particle Moving along a Given Curve Where It Is Required to Find Rate of Change in Its 𝑦-Coordinate Mathematics • Higher Education

A particle is moving along the curve 6𝑦² + 2𝑥² − 2𝑥 + 5𝑦 − 13 = 0. If the rate of change of its 𝑥-coordinate with respect to time as it passes through the point (−1, 3) is 2, find the rate of change of its 𝑦-coordinate with respect to time at the same point.

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Video Transcript

A particle is moving along the curve six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13 equals zero. If the rate of change of its 𝑥-coordinate with respect to time as it passes through the point negative one, three is two, find the rate of change of its 𝑦-coordinate with respect to time at the same point.

So let’s begin by looking at what the question actually wants us to find. It wants us to find the rate of change of its 𝑦-coordinate. Now we recall that the rate of change of a quantity is its derivative with respect to time. So in this case, we’re looking to find d𝑦 by d𝑡. And in fact, we’re given information about the rate of change of the 𝑥-coordinate with respect to time. We’re given information about d𝑥 by d𝑡. At a given point, at the point negative one, three, the value of d𝑥 by d𝑡 is equal to two. And so what we’re going to do first is begin by differentiating the entire equation of our curve. We’ll need to use implicit differentiation to do so.

Remember, we’re differentiating with respect to 𝑡. And the derivative of zero is, of course, zero. We can then go through and differentiate term by term. And of course, we’re going to use implicit differentiation to do so. Let’s begin with six 𝑦 squared. We can differentiate six 𝑦 squared with respect to 𝑦. It will be two times six 𝑦. That simplifies to 12𝑦. And so that means the derivative of this expression with respect to 𝑡 will be 12𝑦 times d𝑦 by d𝑡. We perform a similar process for two 𝑥 squared. We differentiate it with respect to 𝑥. And that’s two times two 𝑥 or four 𝑥. And then we multiply that by d𝑥 by d𝑡.

Next, we differentiate negative two 𝑥 with respect to 𝑡. And to do so, we differentiate with respect to 𝑥 and then times that by d𝑥 by d𝑡. In a similar way, the derivative of five 𝑦 with respect to 𝑡 is five d𝑦 by d𝑡. Then the derivative of our constant, negative 13, is simply equal to zero. We’re now going to factor for d𝑦 by d𝑡 and d𝑥 by d𝑡 just to neat some things up a little bit. When we do, we find that d𝑦 by d𝑡 times 12𝑦 plus five plus d𝑥 by d𝑡 times four 𝑥 minus two is equal to zero.

Since we’re trying to find the value of d𝑦 by d𝑡 at a given point, we’re going to rearrange to make d𝑦 by d𝑡 the subject. We begin by subtracting the term containing d𝑥 by d𝑡 from both sides; then we divide through by 12𝑦 plus five. And so our expression for d𝑦 by d𝑡 is as shown. All that’s left is to perform a few substitutions. Firstly, we know the value of d𝑥 by d𝑡 is equal to two. And we’re interested in finding the rate of change of the 𝑦-coordinate with respect to time at the same point, when 𝑥 is equal to negative one and 𝑦 is equal to three.

And so we substitute all of these values into our expression for d𝑦 by d𝑡. We get negative two times four times negative one minus two over 12 times three plus five. And that gives us a value of 12 over 41. We aren’t given any units for the rate of change of the 𝑥-coordinate. Although, we might assume that it’s units per time. And so the rate of change of the 𝑦-coordinate with respect to time at the same point is 12 over 41 or 12 over 41 units per time.