### Video Transcript

A curve is defined by the parametric equations π₯ is equal to π‘ to the fourth power minus two π‘ squared plus one and π¦ is equal to π‘ to the fifth power. Which of the following is the equation of the line tangent to the graph at nine, 32? Option a, three π¦ minus 10π₯ is equal to six. Option b, 10π¦ minus three π₯ is equal to negative six. Option c, 10π¦ minus three π₯ is equal to 293. Or option d, three π₯ minus 10π¦ is equal to negative 293.

The question gives us a pair of parametric equations. And the question wants us to find the equation of the line tangent to the curve defined by this pair of parametric equations at the point nine, 32. We recall the equation of a line with slope π that passes through the point π₯ one, π¦ one is given by π¦ minus π¦ one is equal to π times π₯ minus π₯ one. In other words, to find the equation of our tangent line, we want to find the slope of our tangent line and the point that our tangent line passes through.

Since weβre trying to find the tangent to the curve at the point nine, 32, our tangent line will pass through the point nine, 32. So we can set π₯ one equal to nine and π¦ one equal to 32. All we need to do now is find the slope of our tangent line. To find the slope of our tangent line, weβd want to find the slope of our curve dπ¦ by dπ₯. However, weβre given π₯ and π¦ as parametric equations. Weβre given π₯ in terms of π‘ and π¦ in terms of π‘.

To get around this, we can recall the following fact about parametric equations which we get by applying a variant of the chain rule. Which tells us if π₯ is some function π of π‘ and π¦ is some function π of π‘, then we can find the slope dπ¦ by dπ₯ by first finding dπ¦ by dπ‘ and then dividing this by dπ₯ by dπ‘. And we can find both of these derivatives by using the power rule for differentiation. First, dπ¦ by dπ‘ is the derivative of π‘ to the fifth power with respect to π‘. And to differentiate this, we multiply by the exponent and then reduce the exponent by one. This gives us five π‘ to the fourth power.

We can then do the same to find dπ₯ by dπ‘; thatβs the derivative of π‘ to the fourth power minus two π‘ squared plus one with respect to π‘. Again, we differentiate each term by using the power rule for differentiation. This gives us four π‘ cubed minus four π‘. Since dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘, we have our slope dπ¦ by dπ₯ is equal to five π‘ to the fourth power divided by four π‘ cubed minus four π‘. So we now have an equation for the slope of our curve in terms of π‘.

And remember, we want to find the slope of the tangent to the curve at the point nine, 32. Thatβs when π₯ is equal to nine and π¦ is equal to 32. So to find the slope of our tangent at this point, we need to find the value of π‘ which gives us π₯ is equal to nine and which gives us π¦ is equal to 32. Weβll start by finding the values of π‘ which give us π¦ is equal to 32.

Substituting π¦ is equal to 32, we get 32 is equal to π‘ to the fifth power. And we can solve this by taking the fifth root of both sides of our equation. We have the fifth root of 32 is equal to two and the fifth root of π‘ to the power of five is equal to π‘. So we can see when π‘ is equal to two, weβll have π₯ is equal to two to the fourth power minus two times two squared plus one and π¦ is equal to two to the fifth power. And we can evaluate both of these expressions to see that π₯ is equal to nine and π¦ is equal to 32.

So when π‘ is equal to two, our parametric curve is at the point nine, 32. This means we can find the slope of our tangent at this point by substituting π‘ is equal to two into our equation for dπ¦ by dπ₯. Substituting π‘ is equal to two into our equation for dπ¦ by dπ₯ gives us five times two to the fourth power divided by four times two cubed minus four times two. We have that two to the fourth power is equal to 16. We have that two cubed is equal to eight and four times two is also equal to eight. Evaluating and simplifying the expressions in our numerator and our denominator, we get 80 divided by 24.

Finally, by canceling the shared factor of eight in our numerator and our denominator, we get 10 divided by three. So weβve shown that the slope of the tangent to the curve at the point nine, 32 is equal to 10 divided by three. So we now have the slope of our tangent line and a point that our line passes through. Substituting our values of π₯ one, π¦ one, and π into our equation for our line gives us π¦ minus 32 is equal to 10 over three times π₯ minus nine.

We could leave our answer like this. However, we can simplify by multiplying both sides of our equation through by three. Multiplying the left-hand side of our equation through by three, we get three π¦ minus 32 times three, which is equal to 96. And when we multiply the right-hand side of our equation through by three, weβll get ten-thirds multiplied by three multiplied by π₯ minus nine. The ten-thirds and the three cancel to give us 10.

We can simplify this equation further by expanding our parentheses. This gives us 10π₯ minus 90. Finally, by subtracting 10π₯ from both sides of the equation and adding 96 to both sides of the equation, we get three π¦ minus 10π₯ is equal to six. Therefore, weβve shown the equation of the tangent line to the curve defined by the pair of parametric equations π₯ is equal to π‘ to the fourth power minus two π‘ squared plus one and π¦ is equal to π‘ to the fifth power at the point nine, 32 is given by option a. Three π¦ minus 10π₯ is equal to six.