Question Video: Finding the Value of the Parameter That Makes a Curve Defined by Parametric Equations Have a Horizontal Tangent Mathematics • Higher Education

A curve has parametric equations π‘₯ = 7π‘šΒ³ + 5π‘šΒ² + π‘š + 4 and 𝑦 = 6π‘šΒ² βˆ’ 6π‘š βˆ’ 8. Find π‘š for which the tangent is horizontal.

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Video Transcript

A curve has parametric equations π‘₯ is equal to seven π‘š cubed plus five π‘š squared plus π‘š plus four and 𝑦 is equal to six π‘š squared minus six π‘š minus eight. Find π‘š for which the tangent is horizontal.

In this question, we are given parametric equations where our coordinates π‘₯ and 𝑦 are given in terms of π‘š. We know that when the tangent is horizontal, its gradient or slope, d𝑦 by dπ‘₯, must be equal to zero. We also know that d𝑦 by dπ‘₯ is equal to d𝑦 by dπ‘š divided by dπ‘₯ by dπ‘š. This can also be written using the chain rule as d𝑦 by dπ‘š multiplied by dπ‘š by dπ‘₯. We multiply d𝑦 by dπ‘š by the reciprocal of dπ‘₯ by dπ‘š.

Let’s begin by considering π‘₯, which is equal to seven π‘š cubed plus five π‘š squared plus π‘š plus four. We can differentiate this term by term to get an expression for dπ‘₯ by dπ‘š. Differentiating seven π‘š cubed gives us 21π‘š squared, differentiating five π‘š squared gives us 10π‘š, and differentiating π‘š gives us one. When we differentiate the constant four, we get zero. Therefore, dπ‘₯ by dπ‘š is equal to 21π‘š squared plus 10π‘š plus one.

We can repeat this process for 𝑦, which is equal to six π‘š squared minus six π‘š minus eight. d𝑦 by dπ‘š is equal to 12π‘š minus six. We can then find an expression for d𝑦 by dπ‘₯ by dividing d𝑦 by dπ‘š by dπ‘₯ by dπ‘š. d𝑦 by dπ‘₯ is equal to 12π‘š minus six divided by 21π‘š squared plus 10π‘š plus one.

For the tangent to be horizontal, we know that d𝑦 by dπ‘₯ must equal zero. This means that we need to solve this equation equal to zero. As the denominator cannot be equal to zero, the numerator must be. 12π‘š minus six must equal zero. Adding six to both sides gives us 12π‘š is equal to six. Dividing both sides of this equation by 12 gives us π‘š is equal to six over 12 or six twelfths. This fraction is equivalent to one-half. The value of π‘š for which the tangent is horizontal is one-half.

Whilst we don’t need to in this question, we could then substitute this value for π‘š back into our equations for π‘₯ and 𝑦. This would give us the coordinate of the point at which the tangent is horizontal.

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