Science:Math Exam Resources/Courses/MATH104/December 2016/Question 07 (a)

MATH104 December 2016

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Question 07 (a)
Consider the equation $(x^{2}+y^{2}-1)^{3}=x^{2}y^{3}$ with the following graph: Find the slope of the tangent line 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?

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Hint
Using the implicit differentiation, find $y'={\frac {dy}{dx}}$.

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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. Note that the slope of tangent line at the point $(1,1)$ is $y'{\bigg |}_{(x,y)=(1,1)}$. To find it, we use implicit differentiation: we consider $y$ as the function of $x$ and differentiate both side of the equation $((x^{2}+y^{2}-1)^{3})'=(x^{2}y^{3})'$ By the chain rule, the left hand side becomes $((x^{2}+y^{2}-1)^{3})'=3(x^{2}+y^{2}-1)^{2}(x^{2}+y^{2}-1)'=3(x^{2}+y^{2}-1)^{2}(2x+2yy').$ On the other hand, using the product rule, we can write the right hand side as $(x^{2}y^{3})'=(x^{2})'y^{3}+x^{2}(y^{3})'=2xy^{3}+x^{2}\cdot 3y^{2}y'=2xy^{3}+3x^{2}y^{2}y'$. Combining both, we get $3(x^{2}+y^{2}-1)^{2}(2x+2yy')=2xy^{3}+3x^{2}y^{2}y'$. Now, we plug $(x,y)=(1,1)$ and simplify the equation. At $(x,y)=(1,1)$, we have $3(1^{2}+1^{2}-1)^{2}\left(2\cdot 1+2\cdot 1y'{\bigg |}_{(x,y)=(1,1)}\right)=2\cdot 1\cdot 1^{3}+3\cdot 1^{2}\cdot 1^{2}y'{\bigg |}_{(x,y)=(1,1)}$ and hence $3\left(2+2y'{\bigg |}_{(x,y)=(1,1)}\right)=2+3y'{\bigg |}_{(x,y)=(1,1)}\implies 6+6y'{\bigg |}_{(x,y)=(1,1)}=2+3y'{\bigg |}_{(x,y)=(1,1)}$ Solving this for $y'$, we get $y'{\bigg |}_{(x,y)=(1,1)}=-{\frac {4}{3}}$. Answer: $\color {blue}-{\frac {4}{3}}$.