A curve has parametric equations 𝑥 is equal to seven 𝑚 cubed plus five 𝑚 squared plus 𝑚 plus four and 𝑦 is equal to six 𝑚 squared minus six 𝑚 minus eight. Find 𝑚 for which the tangent is horizontal.
In this question, we are given parametric equations where our coordinates 𝑥 and 𝑦 are given in terms of 𝑚. We know that when the tangent is horizontal, its gradient or slope, d𝑦 by d𝑥, must be equal to zero. We also know that d𝑦 by d𝑥 is equal to d𝑦 by d𝑚 divided by d𝑥 by d𝑚. This can also be written using the chain rule as d𝑦 by d𝑚 multiplied by d𝑚 by d𝑥. We multiply d𝑦 by d𝑚 by the reciprocal of d𝑥 by d𝑚.
Let’s begin by considering 𝑥, which is equal to seven 𝑚 cubed plus five 𝑚 squared plus 𝑚 plus four. We can differentiate this term by term to get an expression for d𝑥 by d𝑚. Differentiating seven 𝑚 cubed gives us 21𝑚 squared, differentiating five 𝑚 squared gives us 10𝑚, and differentiating 𝑚 gives us one. When we differentiate the constant four, we get zero. Therefore, d𝑥 by d𝑚 is equal to 21𝑚 squared plus 10𝑚 plus one.
We can repeat this process for 𝑦, which is equal to six 𝑚 squared minus six 𝑚 minus eight. d𝑦 by d𝑚 is equal to 12𝑚 minus six. We can then find an expression for d𝑦 by d𝑥 by dividing d𝑦 by d𝑚 by d𝑥 by d𝑚. d𝑦 by d𝑥 is equal to 12𝑚 minus six divided by 21𝑚 squared plus 10𝑚 plus one.
For the tangent to be horizontal, we know that d𝑦 by d𝑥 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. 12𝑚 minus six must equal zero. Adding six to both sides gives us 12𝑚 is equal to six. Dividing both sides of this equation by 12 gives us 𝑚 is equal to six over 12 or six twelfths. This fraction is equivalent to one-half. The value of 𝑚 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 𝑚 back into our equations for 𝑥 and 𝑦. This would give us the coordinate of the point at which the tangent is horizontal.