A bug is moving along the right side of the parabola y=x² at a...

Question

A bug is moving along the right side of the parabola y=x² at a rate such that its distance from the origin is increasing at 1 cm/min.

b. Use the equation y=x² to find an equation relating dy/dt to dx/dt.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a particle is traveling along the curve defined by Y is equal to Xcu. If the particle's distance from the origin is increasing at a constant rate of 2 centimeters per minute, find the relationship between the derivative of Y with respect to T and derivative of X with respect to T. So, we need to determine the relationship between the derivative of Y with respect to T and the derivative of X with respect to T. For this particle that is traveling away from the origin. So, we first need to define the distance away from the origin, and we can use Pythagorean's theorem to do that. So we'll call the distance that the particle is away from the origin as S. and by Pythagorean's theorem, this is going to be equal to the square root of the quantity of X2 plus Y2. So we're looking for a relationship between the derivative of Y with respect with respect to T and the derivative of X with respect to T. So we need to take a derivative with respect to T of both sides of this equation. In doing so, on the left hand side, we'll have the derivative of S with respect to T. And this is going to be equal to. On the right hand side, we'll need to take the derivative with respect to T of the square root of the quantity of X2 plus Y2. So taking the derivative of the square root function, we will get 1 divided by the quantity of 2 multiplied by the square root of the quantity of X2 plus Y2 in quantity. And we'll still have to apply our chain rules, so we need to take the derivative with respect to T of the quantity of X2 plus Y2. So we'll multiply this fraction by the quantity. I'm going to take the derivative with respect to T of X2 + Y2. I get 2 X multiplied by the derivative of X with respect to T. Plus to Y multiplied by the derivative of Y with respect to T. So simplifying this expression, we will get the derivative of S with respect to T. is equal to X multiplied by the derivative of X with respect to T. Plus Y multiplied by the derivative of Y with respect to T. That whole quantity is divided by the square root. Of the quantity of X2 plus Y2. So now what we can do is substitute in our expression for Y, so since Y is equal to X cube, for Y we will substitute N X cubed. We're also told that the um distance that the particle is from the origin is increasing at a constant rate of 2 centimeters per minute. That means that the derivative of S with respect to T is equal to 2, so we can substitute in that value as well. So doing so, we'll have 2 is equal to x multiplied by the derivative of X with respect to T. Plus x cubed multiplied by the derivative of Y with respect to T. That whole quantity is divided by. The square root of the quantity of X2 plus X raised to the 6th power. In quantity. So now we can just simplify this um fraction a bit. So we'll have 2. is equal to, and in the numerator I can factor out an X, so I'll have X, multiplying the quantity of the derivative of X with respect to T, plus X2 multiplied by the derivative of Y with respect to T. In quantity divided by the quantity, and here in the denominator we can factor an X2d out of both terms underneath the square root sign. And when I take the square root of X2, I get the absolute value of X. So I'll have the absolute value of X multiplied by the square root of 1 plus x to the 4th. So now what we can do is multiply both sides of this equation by x as well as the square root of the quantity of 1 plus x to the 4th in quantity and divide by X. In doing so, we will get 2 multiplied by the absolute value of X multiplied by the square root of the quantity of 1 plus x 4th in quantity, divided by X. is equal to the derivative of X with respect to T. Plus x squared. Multiplied by the derivative of Y with respect to T. So this is actually the answer for this problem. So here we actually have a relationship between the derivative of X with respect to T and the derivative of Y with respect to T. So I hope that this explanation has been useful, and I'll see you in the next video.

A bug is moving along the right side of the parabola y=x² at a rate such that its distance from the origin is increasing at 1 cm/min.
b. Use the equation y=x² to find an equation relating dy/dt to dx/dt.

Key Concepts

Related Rates

Implicit Differentiation

Distance Formula

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