Question Video: Finding the First Derivative of a Function Defined by Parametric Equations at a Given Value for the Parameter | Nagwa Question Video: Finding the First Derivative of a Function Defined by Parametric Equations at a Given Value for the Parameter | Nagwa

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Question Video: Finding the First Derivative of a Function Defined by Parametric Equations at a Given Value for the Parameter Mathematics • Third Year of Secondary School

Determine d𝑦/d𝑥 at 𝑡 = 0, given that 𝑥 = (𝑡 − 2)(4𝑡 + 3) and 𝑦 = (3𝑡² − 4)(𝑡 − 3).

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Video Transcript

Determine d𝑦 by d𝑥 at 𝑡 equals zero, given that 𝑥 is equal to 𝑡 minus two multiplied by four 𝑡 plus three and 𝑦 is equal to three 𝑡 squared minus four multiplied by 𝑡 minus three.

In this question, we have a parametric equation where the coordinates 𝑥 and 𝑦 are given in terms of 𝑡. We recall that d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡. This can also be written as d𝑦 by d𝑡 multiplied by d𝑡 by d𝑥, where d𝑡 by d𝑥 is the reciprocal of d𝑥 by d𝑡. Let’s begin by considering our 𝑥-coordinate, 𝑡 minus two multiplied by four 𝑡 plus three. We can distribute the parentheses or expand the brackets here by using the FOIL method.

Multiplying the first terms gives us four 𝑡 squared, multiplying the outside terms gives us three 𝑡, and multiplying the inside terms gives us negative eight 𝑡. Finally, multiplying the last terms gives us negative six. This can be simplified so that 𝑥 is equal to four 𝑡 squared minus five 𝑡 minus six. We can work out an expression for d𝑥 by d𝑡 by differentiating this with respect to 𝑡 term by term. Differentiating four 𝑡 squared gives us eight 𝑡, differentiating negative five 𝑡 gives us negative five, and differentiating any constant gives us zero. Therefore, d𝑥 by d𝑡 is equal to eight 𝑡 minus five.

We can then repeat this process for our 𝑦-coordinate. We know that 𝑦 is equal to three 𝑡 squared minus four multiplied by 𝑡 minus three. Distributing the parentheses here gives us 𝑦 is equal to three 𝑡 cubed minus nine 𝑡 squared minus four 𝑡 plus 12. Once again, we can differentiate this term by term with respect to 𝑡. d𝑦 by d𝑡 is equal to nine 𝑡 squared minus 18𝑡 minus four.

As d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡, this is equal to nine 𝑡 squared minus 18𝑡 minus four divided by eight 𝑡 minus five. We want to calculate this when 𝑡 is equal to zero. Nine 𝑡 squared, 18𝑡, and eight 𝑡 will all be equal to zero. This means that d𝑦 by d𝑥 is equal to negative four over negative five. Dividing a negative number by a negative number gives us a positive answer. Therefore, d𝑦 by d𝑥 is equal to four-fifths.

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