Science:Math Exam Resources/Courses/MATH110/April 2019/Question 04 (b)

MATH110 April 2019

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Question 04 (b)
Suppose the equation given above [$4x^{3}+x(y-3)+(y-3)^{3}=32$] describes a curve in the $xy$-plane. Find the equation of the tangent line to such curve at the point $x=2$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The tangent line at a point $(x_{0},y_{0})$ in the curve has equation $y-y_{0}={\frac {dy}{dx}}(x_{0},y_{0})(x-x_{0}).$ In this problem $x_{0}=2$ and the value of $y_{0}$ is to be determined using the equation of the curve. After this just substitute these values in the equation for ${\frac {dy}{dx}}$ computed in part (a).

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will first calculate the value of $y_{0}$ as in the hint. Since the point $(x_{0},y_{0})$ is a point on the curve, it must satisfy the equation of the curve. Therefore (since we have $x_{0}=2$), $y_{0}$ must satisfy $4(2)^{3}+2(y_{0}-3)+(y_{0}-3)^{3}=32.$ Expanding this the equation becomes $2(y_{0}-3)+(y_{0}-3)^{3}=(y_{0}-3)(2+(y_{0}-3)^{2})=0.$ This has a solution when either $y_{0}-3=0$ or $2+(y_{0}-3)^{2}=0$. The former case yields $y_{0}=3$, while the latter is not possible because the left hand side is always strictly positive. Therefore, the unique point in the curve having $x=2$ is the point $(2,3)$. The slope of the line is obtained by evaluating the formula for $y'$ obtained in part (a) at this point: ${\frac {dy}{dx}}(2,3)={\frac {3-(3)-12(2)^{2}}{(2)+3((3)-3)^{2}}}=-24.$ By substituting these values in the tangent line equation given in the hint, we obtain the equation $y-3=-24(x-2).$ Rearranging this, we conclude that the equation of the tangent line to the curve at $x=2$ is $\color {blue}y=-24x+51$.