Question Video: Finding the Equation of the Tangent to a Curve of a Parametric Equation Involving a Trigonometric Function Mathematics • Third Year of Secondary School

Video Transcript

Find an equation of the tangent to the curve 𝑥 is equal to sin of 𝜋 times 𝑡 and 𝑦 is equal to 𝑡 squared plus 𝑡 at the point zero, two.

The question gives us a pair of parametric equations. And it wants us to find an equation of the tangent to the curve represented by this pair of parametric equations at the point zero, two. To find an equation of a line, we need to find the slope of that line. Since we want our line to be a tangent to the curve, we can find the slope by using d𝑦 by d𝑥. However, we’re given 𝑥 and 𝑦 as a pair of parametric equations.

We recall if 𝑥 is a function of 𝑡 and 𝑦 is a function of 𝑡, then we can find d𝑦 by d𝑥 by dividing the derivative of 𝑦 with respect to 𝑡 by the derivative of 𝑥 with respect to 𝑡. So we set 𝑓 of 𝑡 to be sin of 𝜋𝑡 and 𝑔 of 𝑡 to be 𝑡 squared plus 𝑡. Then by differentiating these with respect to 𝑡, we can find an expression for d𝑦 by d𝑥 . We start by calculating d𝑦 by d𝑡. That’s the derivative of 𝑡 squared plus 𝑡 with respect to 𝑡.

We can differentiate this using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. This gives us two 𝑡 plus one. We can do the same to find d𝑥 by d𝑡. It’s the derivative of sin of 𝜋𝑡 with respect to 𝑡. To differentiate this, we recall that, for any constant 𝑎, the derivative of the sin of 𝑎𝑡 with respect to 𝑡 is equal to 𝑎 times the cos of 𝑎𝑡. Therefore, since 𝜋 is just a constant, the derivative of the sin of 𝜋𝑡 with respect to 𝑡 is just equal to 𝜋 times the cos of 𝜋𝑡.

Now that we found d𝑦 by d𝑡 and d𝑥 by d𝑡, we can find an expression for d𝑦 by d𝑥. It’s equal to two 𝑡 plus one divided by 𝜋 times the cos of 𝜋𝑡. We see that our expression for d𝑦 by d𝑥 is in terms of 𝑡. This means it tells us the slope of the tangent to the curve at the point 𝑡. However, we want to find the slope of the tangent at the point zero, two. So we need to find the value of 𝑡 which gives us the point zero, two. The curve being at the point zero, two is the same as saying its 𝑥-coordinate is equal to zero and its 𝑦-coordinate is equal to two.

To find the values of 𝑡 where our curve is at the point zero, two, we’ll set our 𝑦-value equal to two, giving us the equation two is equal to 𝑡 squared plus 𝑡. We then subtract two from both sides of this equation, giving us that zero is equal to 𝑡 squared plus 𝑡 minus two. And we see in this case we can fully factor the quadratic. It’s equal to 𝑡 plus two multiplied by 𝑡 minus one.

Since we want this to be equal to zero, one of the factors must be equal to zero. Solving these gives us 𝑡 is equal to negative two or 𝑡 is equal to one. We need to check that these values of 𝑡 give us an 𝑥-coordinate of zero. We see when 𝑡 is equal to one, 𝑥 is equal to sin of 𝜋 times one, which is just equal to zero. And we also see when 𝑡 is equal to negative two, 𝑥 is equal to the sin of negative two 𝜋, which is also equal to zero. So we have two different values of 𝑡 at which our parametric curve is at the point zero, two.

At first, this might be confusing. We have several different values of 𝑡. We don’t know which one we should pick. Sometimes our values of 𝑡 are restricted. However, that’s not given in this case. In this case, we’re asked to find an equation of the tangent. So it doesn’t actually matter which value of 𝑡 we pick. We will always get an equation for the tangent. However, it’s worth noting at this point that the value of 𝑡 that you choose could possibly give completely different equations for the tangent.

For example, consider a parametric curve which gives a figure-eight pattern as shown in the sketch. Then, if we were asked to find a tangent at the origin to this curve, we can see that there are two possibilities. Both lines share a slope with different parts of the curve at the origin. In fact, both of these are tangent lines to the curve at the origin. And which tangent line you get would depend on which value of the parameterization variable you choose.

In our case, since we’re just asked to find an equation of the tangent, it doesn’t matter which value of 𝑡 we choose. Let’s start by checking 𝑡 is equal to one. We have the slope of our curve when 𝑡 is equal to one is equal to two times one plus one divided by 𝜋 times the cos of 𝜋 times one, which we can calculate to give us negative three divided by 𝜋.

To be thorough, let’s check what the value of our slope would’ve been if we had chosen 𝑡 is equal to negative two instead. We would’ve gotten a slope of two times negative two plus one all divided by 𝜋 times the cos of 𝜋 times negative two. And we would’ve gotten, again, the slope to be equal to negative three divided by 𝜋. So in this case, we can see the slope is the same regardless of which value of 𝑡 we would’ve gotten. This means our equation of the tangent will be the same in both cases.

To help us find the equation of our tangent line to the curve at the point zero, two, we recall the equation of a line which passes through the point 𝑥 one, 𝑦 one with a slope of 𝑚 is given by 𝑦 minus 𝑦 one is equal to 𝑚 times 𝑥 minus 𝑥 one.

We already found slope of our tangent line. It’s equal to negative three divided by 𝜋. And we already know our tangent line passes through the point zero, two. So we’ll set 𝑥 one equal to zero and 𝑦 one equal to two. This gives us 𝑦 minus two is equal to negative three over 𝜋 times 𝑥 minus zero. Simplifying this and rearranging to make 𝑦 the subject gives us 𝑦 is equal to negative three over 𝜋 times 𝑥 plus two.

Therefore, we’ve shown that an equation of the tangent to the curve 𝑥 is equal to the sin of 𝜋𝑡 and 𝑦 is equal to 𝑡 squared plus two at the point zero, two is 𝑦 is equal to negative three over 𝜋 𝑥 plus two.