Science:Math Exam Resources/Courses/MATH104/December 2015/Question 04 (b)

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MATH104 December 2015

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Question 04 (b)
Determine an equation of the line tangent to the curve $x^{4}-x^{2}y+y^{4}=1$ at the point $(-1,1).$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Clearly the tangent line must pass through $(-1,1),$ where it 'touches' the curve. What is the slope of the line?

Hint 2
The slope of the tangent line we seek is $\left.{\tfrac {dy}{dx}}\right\|_{(x,y)=(-1,1)}.$ However, we are not given an explicit expression for ${\tfrac {dy}{dx}}.$ How might we find ${\tfrac {dy}{dx}}$ from the implicit equation given?

Hint 3
Use implicit differentiation to find ${\tfrac {dy}{dx}}$ , keeping in mind that $y$ is a function of $x$ .

Hint 4
Don't forget to apply the chain rule where necessary – e.g., ${\frac {d}{dx}}(y^{4})=4y^{3}\cdot {\frac {dy}{dx}}.$ (Why? If we let $u(y)=y^{4}$ , then ${\frac {d}{dx}}(y^{4})={\frac {du}{dx}}={\frac {du}{dy}}\cdot {\frac {dy}{dx}}=4y^{3}\cdot {\frac {dy}{dx}}$ as above.)

Checking a solution serves two purposes: helping you if, after having used all the hints, 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.
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Solution
Following the hints, we differentiate the given implicit equation (with respect to $x$ ). ${\begin{aligned}{\frac {d}{dx}}(x^{4}-x^{2}y+y^{4})&={\frac {d}{dx}}(1)\\{\frac {d}{dx}}(x^{4})-{\frac {d}{dx}}({\color {OliveGreen}x^{2}}{\color {OrangeRed}y})+{\frac {d}{dx}}(y^{4})&=0\\4x^{3}-\left({\color {OliveGreen}x^{2}}\cdot {\frac {dy}{dx}}+{\color {OrangeRed}y}\cdot 2x\right)+{\frac {d}{dx}}({\color {Orange}y^{4}})&=0&&{\text{by the product rule }}{\frac {d}{dx}}({\color {OliveGreen}u}{\color {OrangeRed}v})={\color {OliveGreen}u}\cdot {\frac {dv}{dx}}+{\color {OrangeRed}v}\cdot {\frac {du}{dx}}\\4x^{3}-\left(x^{2}\cdot {\frac {dy}{dx}}+2xy\right)+4y^{3}\cdot {\frac {dy}{dx}}&=0&&{\text{by the chain rule }}{\frac {d{\color {Orange}u}}{dx}}={\frac {d{\color {Orange}u}}{dy}}\cdot {\frac {dy}{dx}}\\4x^{3}-2xy+(4y^{3}-x^{2})\cdot {\frac {dy}{dx}}&=0\\{\frac {dy}{dx}}&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\end{aligned}}$ At $(x,y)=(-1,1),{\frac {dy}{dx}}={\frac {2\cdot -1\cdot 1-4\cdot (-1)^{3}}{4\cdot 1^{3}-(-1)^{2}}}={\color {MidnightBlue}{\frac {2}{3}}}.$ The tangent line passes through $(-1,1),$ so the slope-point form of the equation of a line gives $y-1={\color {MidnightBlue}{\tfrac {2}{3}}}(x-(-1)).$ After rearranging, we obtain ${\color {blue}y={\tfrac {2}{3}}(x+1)+1}.$