Science:Math Exam Resources/Courses/MATH104/December 2010/Question 02 (a)

MATH104 December 2010

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Question 02 (a)
Find an equation of the tangent line to the curve $\displaystyle {}x^{3}+xy^{2}+y^{3}=13$ at the point (1,2) on the curve. Please simplify.

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
What is the slope of the tangent line? Do you know one point that the tangent line passes through?

Hint 2
The slope of the tangent line is the derivative of $y$ with respect to $x$, ${\frac {dy}{dx}},$ evaluated at $x=1$ and $y=2.$ Use implicit differentiation to find ${\frac {dy}{dx}}.$

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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. To find the slope of the tangent line, we consider $y$ as a function of $x$ near the point (1,2) and use implicit differentiation to find ${\frac {dy}{dx}}$: $3x^{2}+y^{2}+2xy{\frac {dy}{dx}}+3y^{2}{\frac {dy}{dx}}=0$ Isolating ${\frac {dy}{dx}}$, $\left(2xy+3y^{2}\right){\frac {dy}{dx}}=-3x^{2}-y^{2}.$ Hence, ${\frac {dy}{dx}}={\frac {-3x^{2}-y^{2}}{2xy+3y^{2}}}.$ The slope of the tangent line to the curve at $(x,y)=(1,2)$ is then ${\frac {dy}{dx}}={\frac {-3(1)^{2}-2^{2}}{2(1)(2)+3(2)^{2}}}={\frac {-7}{16}}.$ Since the tangent line passes through the point $(1,2)$ and has slope $-7/16$, its equation is $y-2={\frac {-7}{16}}(x-1).$

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