Lesson Video: Derivatives of Parametric Equations Mathematics

Video Transcript

In this video, we’ll learn how to find the first derivative of a curve defined by parametric equations and find the equations of tangents and normals to these curves. We recall that parametric equations are a set of equations that define a group of quantities as functions of different variables. They appear in kinematics, are often used to describe geometric shapes, and even appear in vector geometry. So it’s important to be able to apply calculus to parametric equations.

Suppose that 𝑓 and 𝑔 are differentiable functions such that 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡 are a pair of parametric equations that describe a curve.

We’ll often need to be able to find the tangent or normal to the curve, unless we need to be able to find the derivative to that curve at a point. There will be times where we can write 𝑦 as a function in terms of 𝑥 and differentiate from there. But it’s not always that simple. And that’s often not the most efficient method anyway.

Instead, we apply the chain rule. In this case, this says that the derivative of 𝑦 with respect to 𝑥 is equal to the derivative of 𝑦 with respect to 𝑡 times the derivative of 𝑡 with respect to 𝑥. But what does that actually mean?

Well, we can see that we’ll be able to differentiate 𝑦 with respect to 𝑡. But what about d𝑡 by d𝑥? Well, we’ll rearrange our function for 𝑥. And we say that 𝑡 is equal to the inverse 𝑓 of 𝑥. Then, as a result of the inverse function theorem, we can say that d𝑡 by d𝑥 is the same as one over d𝑥 by d𝑡. And that leads us to a really nice result. We see that d𝑦 by d𝑥 can be found by multiplying d𝑦 by d𝑡 by one over d𝑥 by d𝑡, or d𝑦 by d𝑥 equals d𝑦 by d𝑡 over d𝑥 by d𝑡. This now means that, given two differentiable functions 𝑦 and 𝑥, we can find the derivative of 𝑦 with respect to 𝑥 by dividing the derivative of 𝑦 with respect to 𝑡 by the derivative of 𝑥 with respect to 𝑡. Let’s now have a look at an example of the application of this formula.

Given that 𝑦 equals negative seven [𝑡] cubed plus eight and 𝑧 equals negative seven [𝑡] squared plus three, find the rate of change of 𝑦 with respect to 𝑧.

When faced with a question about the rate of change of something, we should be thinking about derivatives. Here we want to find the rate of change of 𝑦 with respect to 𝑧. So we’re going to work out d𝑦 by d𝑧. That’s the first derivative of 𝑦 with respect to 𝑧.

We then recall that, given two parametric equations — 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡 — we find d𝑦 by d𝑥 by multiplying d𝑦 by d𝑡 by one over d𝑥 by d𝑡. Or equivalently, by dividing d𝑦 by d𝑡 by d𝑥 by d𝑡. In this example, our two functions are 𝑦 and 𝑧. So we say that d𝑦 by d𝑧 equals d𝑦 by d𝑡 divided by d𝑧 by d𝑡. And we see that we’re going to need to begin by differentiating each function with respect to 𝑡.

We’ll begin by differentiating 𝑦 with respect to 𝑡. Remember, to differentiate a polynomial term, we multiply the term by the exponent and then reduce that exponent by one. So the first derivative of negative 70 cubed is three times negative 70 squared. And actually, the first derivative of eight is zero. Of course, we don’t really need to include that plus zero. So we find that d𝑦 by d𝑡 is equal to negative 21𝑡 squared.

We’ll now repeat this for d𝑧 by d𝑡. This time, the first derivative is two times negative seven 𝑡, which is negative 14𝑡. d𝑦 by d𝑧 is what we get when we divide d𝑦 by d𝑡 by d𝑧 by d𝑡. So that’s negative 21𝑡 squared divided by negative 14𝑡. Of course, a negative divided by a negative is a positive. And we can divide both the numerator and the denominator by 𝑡. Our final step is to simplify by dividing by 21 and 14 by seven. So we find the rate of change of 𝑦 with respect to 𝑧 to be three 𝑡 over two.

Now it’s important to note that, occasionally, we can use substitution to express 𝑦 in terms of 𝑥 and differentiate as normal. For instance, if we had two parametric equations — 𝑦 equals five 𝑡 squared and 𝑥 equals three 𝑡 minus two — we can write our equation for 𝑥 as 𝑡 equals 𝑥 plus two over three. Then we replace 𝑡 with 𝑥 plus two over three. And we find that 𝑦 is equal to five times 𝑥 plus two over three all squared. We could then differentiate from here, not usually efficient as you can see. In our next example, we’ll see how the chain rule method holds for slightly more complicated parametric equations.

Find the derivative of seven 𝑥 plus four sin 𝑥 with respect to cos 𝑥 plus one at 𝑥 equals 𝜋 by six.

Let’s begin by defining our two functions. We’ll let 𝑦 be equal to seven 𝑥 plus four sin 𝑥. And we’ll define cos 𝑥 plus one as 𝑧. We then recall that, given two parametric equations — 𝑥 equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡 — we can find d𝑦 by d𝑥 by multiplying d𝑦 by d𝑡 by one over d𝑥 by d𝑡. Or equivalently, by dividing d𝑦 by d𝑡 by d𝑥 by d𝑡.

Now in this case, our two functions are 𝑦 and 𝑧. And they are in terms of 𝑥. So we can see that d𝑦 by d𝑧 must be equal to d𝑦 by d𝑥 divided by d𝑧 by d𝑥. So we’re going to need to begin by differentiating each of our functions with respect to 𝑥. The first derivative of seven 𝑥 is seven. And when we differentiate sin 𝑥, we get cos 𝑥. So we see that d𝑦 by d𝑥 here is seven plus four cos of 𝑥.

We also know that if we differentiate cos of 𝑥, we get negative sin of 𝑥. So that’s d𝑧 by d𝑥. It’s negative sin 𝑥. d𝑦 by d𝑧 is the quotient. It’s seven plus four cos of 𝑥 divided by negative sin 𝑥. But we’re not quite finished. We’re looking to find the derivative at the point where 𝑥 is equal to 𝜋 by six. So we’re going to substitute 𝑥 for 𝜋 by six in our expression. That’s seven plus four cos of 𝜋 by six over negative sin of 𝜋 by six.

Cos of 𝜋 by six is root three over two. And sin of 𝜋 by six is one-half. When we divide by one-half, that’s the same as multiplying the numerator by two. And so we see that the derivative of our function, seven 𝑥 plus four sin 𝑥, with respect to cos of 𝑥 plus one at 𝑥 equals 𝜋 by six is negative 14 minus four root three.

So far, we’ve established how to find the derivative and evaluate it at a point. We then recall that the derivative evaluated at a point is the gradient of the tangent to the curve at that point. And this means we can use coordinate geometry to find the equation of a tangent to a curve. Let’s see what that looks like.

Find the equation of the tangent to the curve 𝑥 equals five sec 𝜃 and 𝑦 equals five tan 𝜃 at 𝜃 equals 𝜋 by six.

To find the equation of a tangent, we must begin by calculating its gradient. That, of course, is the value of the derivative at that point. So we’re going to need to begin by calculating the value of d𝑦 by d𝑥 when 𝜃 is equal to 𝜋 by six.

Well, of course, these are parametric equations. There’s an equation for 𝑥 in terms of 𝜃 and an equation for 𝑦 in terms of 𝜃. We recall that, given two parametric equations — 𝑥 is some function in 𝜃 and 𝑦 is some other function in 𝜃 — we can find d𝑦 by d𝑥 by dividing d𝑦 by d𝜃 by d𝑥 by d𝜃. So we see we’re going to need to begin by differentiating each of our functions with respect to 𝜃. We’ll use the general result for the derivative of sec of 𝑥 and tan of 𝑥 to do so.

The derivative of sec of 𝑥 is sec 𝑥 tan 𝑥. And the derivative of tan 𝑥 is sec squared 𝑥. This means that d𝑥 by d𝜃 is five sec 𝜃 tan 𝜃. And then d𝑦 by d𝜃 is five sec squared 𝜃. d𝑦 by d𝑥 is the quotient to these. It’s five sec squared 𝜃 divided by five sec 𝜃 tan 𝜃. And of course, we can simplify by dividing through by five and by sec 𝜃. We then recall that sec 𝜃 is the same as one over cos 𝜃. We then recall that sec 𝜃 is the same as one over cos 𝜃. And we’re going to be dividing this by tan 𝜃, or dividing by sin 𝜃 over cos 𝜃.

Now dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So we’re going to be multiplying one over cos 𝜃 by cos 𝜃 over sin 𝜃. And that’s great because we have a little more canceling to do. We’ve now fully simplified the expression for the derivative of 𝑦 with respect 𝑥. It’s one over sin 𝜃. We might write that as csc 𝜃.

Remember, we were looking to find the value of the gradient of the tangent at 𝜃 equals 𝜋 by six. So let’s work out d𝑦 by d𝑥 when 𝜃 is 𝜋 by six. That’s one over sin of 𝜋 by six. And since sin of 𝜋 by six is one-half, we find the gradient of the tangent to be two.

Now that we have the gradient to the curve, let’s clear some space and see what else we need. The equation of a straight line with the gradient of 𝑚 that passes through a point with Cartesian coordinates 𝑥 one, 𝑦 one is 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one. We quite clearly know the value of 𝑚. But we don’t have an 𝑥𝑦-coordinate we can use.

But by substituting 𝜃 equals 𝜋 by six into each of our original equations, we’ll find the coordinate of the point where the tangent meets the curve. We get 𝑥 equals five sec of 𝜋 by six. And actually that’s 10 root three over three. And then we get 𝑦 to be equal to five tan of 𝜋 by six. And that’s five root three over three.

Substituting what we have into our formula for the equation of a straight line, and we get 𝑦 minus five root three over three equals two times 𝑥 minus 10 root three over three. When we distribute the parentheses, we get two 𝑥 minus 20 root three over three on the right-hand side. We rearrange by subtracting two 𝑥 and adding 20 root three over three to both sides to find the equation to be 𝑦 minus two 𝑥 plus 15 root three over three equals zero.

All that’s left is for us to simplify our fraction. And when we do, we find the equation of the tangent to our curve when 𝜃 is equal to 𝜋 by six to be 𝑦 minus two 𝑥 plus five root three equals zero.

We’ll now discuss intervals of increase and decrease for a curve defined by parametric functions.

Suppose we have a pair of parametric equations such that 𝑥 equals 𝑡 squared plus 𝑡 and 𝑦 is equal to two 𝑡 minus one. We might wish to work out the nature of the curve over various intervals. In other words, what are the intervals of increase or decrease of our curve?

Remember, we say a graph that’s described by the equation 𝑦 equals 𝑓 of 𝑥 is increasing when its first derivative is greater than zero. And it’s decreasing when its first derivative is less than zero. So to find intervals of increase or decrease, we usually find an expression for d𝑦 by d𝑥 and then establish over which intervals that’s greater than zero or less than zero. In this case, we end up with d𝑦 by d𝑥 equals two over two 𝑡 plus one.

Now we have a little bit of a problem here. This derivative cannot help us to find intervals of increase or decrease. And that’s because the curve looks a little something like this. Notice that the graph is both increasing and decreasing for all values of 𝑥 greater than zero. What we can do though is consider the derivatives with respect to our parameter 𝑡. For instance, should we choose to work out whether the curve is increasing or decreasing when 𝑡 is equal to negative four, we find d𝑦 by d𝑥 to be equal to negative two-sevenths. It’s less than zero, and therefore it’s decreasing at this point. So we can establish what’s happening at key points. It’s just very difficult to work out intervals of increase or decrease.

In this video, we saw that parametric equations are a set of equations that define a group of quantities as functions of different variables. We saw that, for two parametric equations — 𝑥 equals 𝑓 of 𝑡 and 𝑦 equals 𝑔 of 𝑡 — d𝑦 by d𝑥 is d𝑦 by d𝑡 over d𝑥 by d𝑡. We saw that we can use the derivative to find the nature of the curve at various values of 𝑡. But that we need to be really careful because we can’t always evaluate intervals of increase and decrease.