Question Video: Finding the Value of a Variable That Makes a Curve with Parametric Equations Have a Vertical Tangent | Nagwa Question Video: Finding the Value of a Variable That Makes a Curve with Parametric Equations Have a Vertical Tangent | Nagwa

# Question Video: Finding the Value of a Variable That Makes a Curve with Parametric Equations Have a Vertical Tangent Mathematics • Third Year of Secondary School

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Find the value of π at which the curve π₯ = 8πΒ³ + 5πΒ² + π β 1, π¦ = 5πΒ² β π + 2 has a vertical tangent.

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### Video Transcript

Find the value of π at which the curve π₯ equals eight π cubed plus five π squared plus π minus one, π¦ equals five π squared minus π plus two has a vertical tangent.

Now, whenever weβre thinking about tangents to curves, weβre always thinking about the derivative of the expression that describes that curve. Now, in fact, here our curve is described using parametric equations. We have π₯ in terms of π and π¦ in terms of π. So letβs say π₯ is some function of π, π of π, and π¦ is some function of π, π of π. The derivative of π¦ with respect to π₯ from these parametric equations is dπ¦ by dπ divided by dπ₯ by dπ. So how do we use the derivative of our curve with respect to π₯ to find the location of a vertical tangent? Now, in fact, if dπ¦ by dπ is not equal to zero but dπ₯ by dπ is equal to zero, then there has to be a vertical tangent line. So in other words, we need to find the values of π such that dπ₯ by dπ equals zero.

Letβs begin then by differentiating π₯ with respect to π. We can use the power rule for differentiation to do so. Letβs say weβre differentiating some expression ππ to the πth power, where π and π are real numbers. We multiply by the value of the exponent and then reduce that exponent by one. So the derivative with respect to π is π times π times π to the power of π minus one. So to differentiate π₯ with respect to π, weβll do this but term by term.

The derivative of eight π cubed is three times eight π squared. The derivative of five π squared is two times five π to the first power. The derivative of π is one times π to the zeroth power. And the derivative of a constant is zero since, technically, this is one π to the power of zero. And of course, since π to the power of zero is one, this simplifies to 24π squared plus 10π plus one. And we want dπ₯ by dπ to be equal to zero. So letβs set this equal to zero and solve for π.

Weβll need to factor the right-hand side, so when we do, we get six π plus one times four π plus one. And for the product of these two expressions to be equal to zero, either one or other of the expressions must itself be equal to zero. So six π plus one equals zero or four π plus one equals zero.

To solve each expression for π, weβll subtract one from both sides. So we get six π equals negative one and four π equals negative one, respectively. Then we divide both sides of our first equation by six, giving us π equals negative one-sixth. And we divide both sides of our second equation by four, so we get π equals negative one-quarter. So this implies the existence of a vertical tangent line to the curve at π equals negative one-sixth and π equals negative one-quarter.

We will double-check though that the value of dπ¦ by dπ is not equal to zero at either of these points. So letβs take our expression for π¦ and differentiate with respect to π. The derivative of five π squared with respect to π is 10π, and the derivative of negative π is negative one. Similarly, the derivative of two with respect to π is zero. So dπ¦ by dπ is 10π minus one. Letβs substitute π equals negative one-sixth into this expression and check itβs not equal to zero. In fact, itβs equal to negative eight-thirds.

So we have a vertical tangent line at π equals negative one-sixth. Similarly, when π is equal to negative one-quarter, dπ¦ by dπ gives us a value of negative seven over two. Once again, this is not equal to zero. So we know that there must be a vertical tangent line when π equals negative one-quarter. The value of π at which the curve has a vertical tangent is negative one-sixth and negative one-quarter.

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