### Video Transcript

Find the equation of the normal to the curve π¦ cubed is equal to six π₯ squared minus six π₯ plus one at π₯ is equal to one.

We need to find the equation of the normal line to a curve when π₯ is equal to one. To start, because the normal line is a straight line, this equation will take the form of a straight line. We recall this is of the form π¦ minus π¦ one is equal to π times π₯ minus π₯ one where our line passes through the point π₯ one, π¦ one and has a slope of π. So, to find the equation of this normal line, we need to find two things. First, we need to find a point which our straight line passes through. And we also need to find its slope.

Letβs start by finding a point our line passes through. The question asks us to find the normal to the curve at the point where π₯ is equal to one. This tells us that our normal line has to pass through the point on our curve when π₯ is equal to one. So, we know the value of π₯ one will be one. And we can find the value of π¦ one by substituting π₯ is equal to one into our equation for the curve. Doing this, we get π¦ one cubed is equal to six times one squared minus six times one plus one.

And we can just calculate this. The right-hand side of our equation simplifies to give us one. Then, we can just solve for π¦ one by taking the cube roots of both sides of this equation. We have that π¦ one is the cube root of one, which is of course just equal to one. So, we found the point our normal line passes through. All thatβs left to do now is find the value of its slope.

To do this, we need to recall that the normal line to a curve at a point will be perpendicular to the tangent line of the same curve at the same point. So, the first thing we want to do is find the slope of the tangent line to our curve when π₯ is equal to one. However, we can see we have a problem. Our curve is not given π¦ as some function in π₯. Instead, we have π¦ cubed is some function in π₯.

Luckily, we know the cube function has an inverse for any real value. We can just take the cube root of both sides. We get π¦ is equal to the cube root of six π₯ squared minus six π₯ plus one. And we want to use this to find the slope of the tangent line to our curve. We need to find an expression for dπ¦ by dπ₯. This means we need to differentiate the cube root of six π₯ squared minus six π₯ plus one.

And thereβs a couple of different ways of doing this. For example, we could do this by using the chain rule. However, because our outer function is a power function, we can use the general power rule. Either method will work. Itβs personal preference which you would prefer. Weβre going to do this by using the general power rule.

We recall the general power rule tells us for differentiable function π of π₯ in any real constant π, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ all raised to the power of π minus one. And we can see this applies in this case. Our exponent π will be equal to one-third, and our function π of π₯ will be the quadratic six π₯ squared minus six π₯ plus one.

So, weβll find our expression for dπ¦ by dπ₯ by using the general power rule. We need to multiply by π, which is one-third. Then, we need to differentiate our inner function. We can, of course, do this term by term by using the power rule for differentiation. We want to multiply by our exponent of π₯ and reduce this exponent by one. We get 12π₯ minus six.

Finally, we just multiply this by our original function. However, we reduce the exponent by one. So, we have dπ¦ by dπ₯ is one-third times 12π₯ minus six multiplied by six π₯ squared minus six π₯ plus one all raised to the power of one-third minus one. And we can simplify this expression slightly to get dπ¦ by dπ₯ is equal to four π₯ minus two multiplied by six π₯ squared minus six π₯ plus one all raised to the power of negative two over three.

Now, weβre ready to find the slope of our tangent line to the curve when π₯ is equal to one. We just substitute π₯ is equal to one into our expression for dπ¦ by dπ₯. Doing this, we get four times one minus two multiplied by six times one squared minus six times one plus one all raised to the power of negative two over three. And we can simplify this expression. In our first set of parentheses, four times one minus two is equal to two. Similarly, in our second set of parentheses, six times one squared minus six times one plus one is equal to one.

So, the slope of our tangent line when π₯ is equal to one is given by two times one to the power of negative two over three. But one raised to any exponent is just equal to one. So, this simplifies to give us two times one, which is of course just equal to two. So, we found the slope of our tangent line at this point. But remember, weβre not looking for the slope of our tangent line; weβre looking for the slope of the normal line at this point.

Therefore, our normal line is perpendicular to our line with slope two. And to find the slope of a perpendicular line to this, we just need to take the negative reciprocal. In other words, we need to divide negative one by two to find the slope of our normal line. π is equal to negative one-half.

Now that we know a point that our normal line passes through and its slope, we can substitute these values into the equation for our line to find the equation of our normal to the curve at this point. Substituting π₯ one is equal to one, π¦ one is equal to one, and π is equal to negative a half into our equation for our line, we get π¦ minus one is equal to negative a half multiplied by π₯ minus one. And we can simplify this expression. Weβll distribute negative a half over our parentheses. This gives us π¦ minus one is equal to negative π₯ over two plus one-half.

And weβll do one more thing to simplify this expression. Weβll write all of our terms on the same side of the equation. Doing this and then simplifying, we get π₯ over two plus π¦ minus three over two is equal to zero, which is our final answer. Therefore, we were able to find the equation of the normal to the curve π¦ cubed is equal to six π₯ squared minus six π₯ plus one at π₯ is equal to one. It has the equation π₯ over two plus π¦ minus three over two is equal to zero.