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‘š at which the curve π‘₯ = 8π‘šΒ³ + 5π‘šΒ² + π‘š βˆ’ 1, 𝑦 = 5π‘šΒ² βˆ’ π‘š + 2 has a vertical tangent.

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Video Transcript

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

Now, whenever we’re thinking about tangents to curves, we’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‘š and 𝑦 in terms of π‘š. So let’s say π‘₯ is some function of π‘š, 𝑓 of π‘š, and 𝑦 is some function of π‘š, 𝑔 of π‘š. The derivative of 𝑦 with respect to π‘₯ from these parametric equations is d𝑦 by dπ‘š divided by dπ‘₯ by dπ‘š. 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 d𝑦 by dπ‘š is not equal to zero but dπ‘₯ by dπ‘š 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‘š such that dπ‘₯ by dπ‘š equals zero.

Let’s begin then by differentiating π‘₯ with respect to π‘š. We can use the power rule for differentiation to do so. Let’s say we’re differentiating some expression π‘Žπ‘š to the 𝑛th power, where π‘Ž and 𝑛 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‘š is 𝑛 times π‘Ž times π‘š to the power of 𝑛 minus one. So to differentiate π‘₯ with respect to π‘š, we’ll do this but term by term.

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

We’ll need to factor the right-hand side, so when we do, we get six π‘š plus one times four π‘š 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‘š plus one equals zero or four π‘š plus one equals zero.

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

We will double-check though that the value of d𝑦 by dπ‘š is not equal to zero at either of these points. So let’s take our expression for 𝑦 and differentiate with respect to π‘š. The derivative of five π‘š squared with respect to π‘š is 10π‘š, and the derivative of negative π‘š is negative one. Similarly, the derivative of two with respect to π‘š is zero. So d𝑦 by dπ‘š is 10π‘š minus one. Let’s substitute π‘š equals negative one-sixth into this expression and check it’s not equal to zero. In fact, it’s equal to negative eight-thirds.

So we have a vertical tangent line at π‘š equals negative one-sixth. Similarly, when π‘š is equal to negative one-quarter, d𝑦 by dπ‘š 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‘š equals negative one-quarter. The value of π‘š at which the curve has a vertical tangent is negative one-sixth and negative one-quarter.

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