### Video Transcript

Find the equation of the tangent to the curve π₯ equals π‘ cubed plus one, π¦ equals π‘ to the fourth power plus π‘ at the point corresponding to the value π‘ equals negative one.

Remember, the tangent to a curve is given by a straight line. That line has the same slope as the slope of the curve at the point where the tangent touches. And so if we can find the slope β letβs call that π β of the curve at that given point β here thatβs π‘ equals negative one β and we know the coordinates of the point where the tangent meets the curve, we can find its equation by using the formula π¦ minus π¦ one equals π times π₯ minus π₯ one, where the coordinates π₯ one, π¦ one is the point where the tangent meets the curve.

And so it should be quite clear that weβre going to need to find the slope of the curve at that point, specifically at the point where π‘ equals negative one. But we might notice we have a pair of equations that describe our curve: π₯ equals π‘ cubed plus one and π¦ equals π‘ to the fourth power plus π‘. In fact, this is a pair of parametric equations. Theyβre defined by a third variable π‘. And we can find the slope by finding the derivative at the point where π‘ equals negative one.

Specifically, the formula we can use to find dπ¦ by dπ₯ given a pair of parametric equations in π‘ is dπ¦ by dπ₯ equals dπ¦ by dπ‘ divided by dπ₯ by dπ‘. Now π₯ is given by the equation π‘ cubed plus one. So we can differentiate π₯ with respect to π‘ by using the general power rule. That is, we multiply by the exponent and then reduce that exponent by one. So the derivative with respect to π‘ of π‘ cubed is three π‘ squared. The derivative of a constant is zero, so we found dπ₯ by dπ‘.

Similarly, we have an equation for π¦ in terms of π‘. Itβs π‘ to the fourth power plus π‘. Once again, we can differentiate by using the general power rule. The derivative of π‘ to the fourth power is four π‘ cubed, whilst the derivative of π‘ with respect to π‘ is just one. Then dπ¦ by dπ₯ is the quotient of these. Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘. We can write that as four π‘ cubed plus one over three π‘ squared.

Then remember, the slope, the value of π that weβre trying to find, is the value of the derivative at the point where π‘ equals negative one. So weβre going to substitute π‘ equals negative one into the expression four π‘ cubed plus one over three π‘ squared. That gives us four times negative one cubed plus one over three times negative one squared. We get negative four plus one over three. And thatβs negative three divided by three, which is negative one. And thatβs great because we found the value of the slope of the tangent to the curve when π‘ equals negative one.

We still need to find the point where the tangent meets the curve. So weβre going to take the value of π‘ equals negative one and substitute it into our equation for π₯ and our equation for π¦. When π‘ is negative one, our equation for π₯ becomes negative one cubed plus one. Weβll define it to be π₯ sub one. And itβs the π₯-coordinate of the point where the tangent to the curve meets the curve itself. Now negative one cubed is negative one, and then when we add one, we get zero. Then in a similar way, we can find the value of π¦ sub one by substituting π‘ equals negative one into the expression π‘ to the fourth power plus π‘. Itβs negative one to the fourth power plus negative one, which is, once again, zero.

So in fact, weβre finding the equation of the tangent to the curve at the origin. We substitute π¦ sub one equals zero, π equals negative one, and π₯ sub one equals zero into the equation of a straight line. And we get π¦ minus zero equals negative one times π₯ minus zero. Then distributing the parentheses on the right-hand side, and we simply get negative π₯. And so the equation of the tangent to the curve at the point where π‘ equals negative one is π¦ equals negative π₯.