# Video: Finding the Equation of the Tangent to the Curve Defined by Parametric Equations Involving Cubic and Cube Root Functions

Given that 𝑥 = 2𝑡³ − 9 and 𝑦 = ∛(7𝑡 + 8), determine the equation of the tangent to the curve at 𝑡 = −1.

05:39

### Video Transcript

Given that 𝑥 is equal to two 𝑡 cubed minus nine and 𝑦 is equal to the cube root of seven 𝑡 plus eight, determine the equation of the tangent to the curve at 𝑡 is equal to negative one.

The question gives us a pair of parametric equations. And it wants us to determine the equation of the tangent to the curve at the point where 𝑡 is equal to negative one. We recall we can find the equation of a line with a slope of 𝑚 which passes through the point 𝑥 one, 𝑦 one by using 𝑦 minus 𝑦 one is equal to 𝑚 times 𝑥 minus 𝑥 one.

So to find the equation of the tangent line to the curve, we need to find a point which it passes through and the slope of the tangent. We can find the point that our tangent line passes through by just substituting 𝑡 is equal to negative one into our pair of parametric equations. We also know the slope of the tangent is given by the change in 𝑦 with respect to 𝑥. And we recall, for a pair of parametric equations 𝑥 is equal to 𝑓 of 𝑡 and 𝑦 is equal to 𝑔 of 𝑡, we can find d𝑦 by d𝑥 by dividing the derivative of 𝑦 with respect to 𝑡 by the derivative of 𝑥 with respect to 𝑡.

So we now know how to find the slope of our tangent and a point which our tangent passes through. Let’s start by finding the slope of our tangent. We need to find d𝑦 by d𝑡 and d𝑥 by d𝑡. We have d𝑦 by d𝑡 is equal to the derivative of the cube root of seven 𝑡 plus eight with respect to 𝑡. We see that this is the composition of two functions. So if we let 𝑢 be equal to seven 𝑡 plus eight, we have 𝑦 is equal to the cube root of 𝑢, and 𝑢 is a function of 𝑡. So we can use the chain rule.

The chain rule tells us the derivative of 𝑦 with respect to 𝑡 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑡, if 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑡. We see the cube root of 𝑢 is just equal to 𝑢 to the power of one-third. So we can differentiate this by using the power rule for differentiation. We multiply by the exponent and then reduce the exponent by one. We now need to multiply this by the derivative of 𝑢 with respect to 𝑡. We can differentiate seven 𝑡 plus eight with respect to 𝑡 by using the power rule for differentiation. It’s just equal to seven.

Finally, we can rewrite d𝑦 by d𝑡 in terms of 𝑡 by substituting 𝑢 is equal to seven 𝑡 plus eight. So by using this substitution and rearranging our expression, we have d𝑦 by d𝑡 is equal to seven divided by three times seven 𝑡 plus eight to the power of two over three. We can do the same to find d𝑥 by d𝑡. It’s equal to the derivative of two 𝑡 cubed minus nine with respect to 𝑡. We can evaluate this by using the power rule for differentiation. It’s equal to six 𝑡 squared.

We can use these to find an expression for the slope d𝑦 by d𝑥. It’s equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡 or d𝑦 by d𝑡 multiplied by the reciprocal of d𝑥 by d𝑡. This gives us d𝑦 by d𝑥 is equal to seven divided by three times seven 𝑡 plus eight to the power of two over three multiplied by the reciprocal of six 𝑡 squared.

We can then use this to find the slope of our tangent line when 𝑡 is equal to negative one. We substitute 𝑡 is equal to negative one into this expression for the slope. This gives us a slope of seven divided by three times seven times negative one plus eight to the power of two over three multiplied by the reciprocal of six times negative one squared. We have seven times negative one plus eight is just equal to one. And then one to the power of two-thirds is just equal to one. And then negative one squared is just equal to one. And so the reciprocal of six times one is just one-sixth. So this simplifies to give us the slope of our tangent is seven-thirds times one-sixth which we can calculate to give us seven divided by 18.

So we found the slope of our tangent line. We now need to find the 𝑥- and 𝑦-coordinate of a point our line passes through. To do this, we’ll substitute 𝑡 is equal to negative one into each of our parametric equations. Substituting 𝑡 is equal to negative one into our parametric equation for 𝑥 gives us 𝑥 is equal to two times negative one cubed minus nine, which we can calculate to be negative 11. Substituting 𝑡 is equal to negative one into our parametric equation for 𝑦 gives us 𝑦 is equal to the cube root of seven times negative one plus eight, which is, of course, just equal to one.

So our tangent line passes through the point negative 11, one and has a slope of seven divided by 18. Using our formula for the equation of our line, we have the equation of our tangent when 𝑡 is equal to negative one is given by 𝑦 minus one is equal to seven over 18 multiplied by 𝑥 minus negative 11. We can simplify this equation by multiplying both sides through by 18. Multiplying 𝑦 minus one by 18 gives us 18𝑦 minus 18. And since seven over 18 multiplied by 18 is just equal to seven, we have the right-hand side of our equation is seven multiplied by 𝑥 minus negative 11.

We know that subtracting negative 11 is the same as adding 11. This gives us 18𝑦 minus 18 is equal to seven times 𝑥 plus 11. Multiplying out the parentheses, rearranging, and simplifying, we can show that the equation of our line is given by negative seven 𝑥 plus 18𝑦 minus 95 is equal to zero. Therefore, we’ve shown the equation to the tangent of the curve 𝑥 is equal to two 𝑡 cubed minus nine and 𝑦 is equal to the cube root of seven 𝑡 plus eight when 𝑡 is equal to negative one. Is given by the equation negative seven 𝑥 plus 18𝑦 minus 95 is equal to zero.