Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 09 (a)

MATH 180 December 2017

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Question 09 (a)
The curve $$\left(x^{2}+y^{2}-2x\right)^{2}=x^{2}+y^{2}$$ is known as a Limaçon of Pascal. (a) Find the equation of the line tangent to the Limaçon at the point ${\textstyle (0,1)}$.

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
Use implicit differentiation.

Hint 2
Recall that ${\textstyle y}$ is considered as a function of ${\textstyle x}$. Remember to use the chain rule.

Hint 3
The equation of tangent line at the point ${\textstyle (x,y)=(a,b)}$ is given by ${\textstyle y=b+{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(a,b)}(x-a)}$.

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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. According to Hint 1 and Hint 2, we differentiate both sides with respect to ${\textstyle x}$ using chain rule and power rule: Let $u=x^{2}+y^{2}-2x$. Then, the left hand side of the equation is $u^{2}$. Considering $y$ as a function of $x$, $u$ can be also considered as a function of $x$. Taking ${\dfrac {d}{dx}}$ to the equation, we have $${\dfrac {d(u^{2})}{dx}}=2u\cdot {\dfrac {du}{dx}}=2(x^{2}+y^{2}-2x){\dfrac {d}{dx}}(x^{2}+y^{2}-2x)=2x+2y{\dfrac {dy}{dx}},$$ $$2(x^{2}+y^{2}-2x)\left(2x+2y{\dfrac {dy}{dx}}-2\right)=2x+2y{\dfrac {dy}{dx}}.$$ Then, plugging $(x,y)=(0,1)$, $$2(0^{2}+1^{2}-2\cdot 0)\left(2\cdot 0+2\cdot 1{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)}-2\right)=2\cdot 0+2\cdot 1{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)},$$ which can be simplified as $$2{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)}-2={\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)},$$ $${\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)}=2.$$ This is the slope of the tangent line. Using Hint 3, the equation of the tangent line at ${\textstyle (x,y)=(0,1)}$ is $$y=1+{\dfrac {dy}{dx}}{\bigg |}_{(x,y)=(0,1)}(x-0),$$ or $$y=1+2x.$$ Answer: the equation of the tangent line at the point ${\textstyle (0,1)}$ is ${\textstyle {\color {blue}y=1+2x}}$.