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

MATH 180 December 2017

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[hide] Question 09 (b)
The curve $$\left(x^{2}+y^{2}-2x\right)^{2}=x^{2}+y^{2}$$ is known as a Limaçon of Pascal. (b) Find the coordinates of one point on the Limaçon where ${\textstyle {\frac {dy}{dx}}}$ is undefined. You must justify your answer.

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.

[show] Hint 1
The derivative is undefined if it has a zero denominator.

[show] Hint 2
Use the expression for ${\textstyle {\dfrac {dy}{dx}}}$ from part (a).

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[show] 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. By part (a), we have $$2(x^{2}+y^{2}-2x)\left(2x+2y{\dfrac {dy}{dx}}-2\right)=2x+2y{\dfrac {dy}{dx}}.$$ We solve for ${\textstyle {\dfrac {dy}{dx}}}$: By dividing 2 on both side of the equation and expanding the left hand side, we get $$(x^{2}+y^{2}-2x)\left(2x-2+2y{\dfrac {dy}{dx}}\right)=x+y{\dfrac {dy}{dx}},$$ $$(2x-2)(x^{2}+y^{2}-2x)+2y(x^{2}+y^{2}-2x){\dfrac {dy}{dx}}=x+y{\dfrac {dy}{dx}}.$$ Then, collecting the terms with ${\dfrac {dy}{dx}}$ to the left hand side of the equation, $$2y(x^{2}+y^{2}-2x){\dfrac {dy}{dx}}-y{\dfrac {dy}{dx}}=x-(2x-2)(x^{2}+y^{2}-2x),$$ $$y(2x^{2}+2y^{2}-4x-1){\dfrac {dy}{dx}}=x-2(x-1)(x^{2}+y^{2}-2x),$$ $${\dfrac {dy}{dx}}={\dfrac {x-2(x-1)(x^{2}+y^{2}-2x)}{y(2x^{2}+2y^{2}-4x-1)}}.$$ Therefore, if ${\textstyle (x,y)}$ is a point on the Limaçon, then $${\dfrac {dy}{dx}}={\dfrac {x-2(x-1)(x^{2}+y^{2}-2x)}{y(2x^{2}+2y^{2}-4x-1)}}.$$ By Hint 2, in order that ${\textstyle {\dfrac {dy}{dx}}}$ is undefined, we need to find a point where the denominator of the above expression is zero. Clearly, this happens when ${\textstyle y=0}$. However, we are asked to find the coordinates, meaning that we must find the value of ${\textstyle x}$ when ${\textstyle y=0}$. To do this, we put ${\textstyle y=0}$ into the implicit equation ${\textstyle (x^{2}+y^{2}-2x)^{2}=x^{2}+y^{2}=0}$ to get $$(x^{2}-2x)^{2}=x^{2}.$$ Using the factorization for the difference of two squares, namely ${\textstyle a^{2}-b^{2}=(a-b)(a+b)}$, we can solve the above equation as follows: $$(x^{2}-2x)^{2}-x^{2}=0,$$ $$(x^{2}-2x-x)(x^{2}-2x+x)=0,$$ $$(x^{2}-3x)(x^{2}-x)=0,$$ $$x(x-3)\cdot x(x-1)=0,$$ $$x^{2}(x-1)(x-3)=0.$$ So ${\textstyle x=0}$, ${\textstyle x=1}$ or ${\textstyle x=3}$. We have found three points ${\textstyle (x,y)=(0,0)}$, ${\textstyle (x,y)=(1,0)}$ and ${\textstyle (x,y)=(3,0)}$. Since we are asked to find one, we may just pick, say, ${\textstyle (x,y)=(0,0)}$. Answer: a point on the Limaçon where ${\textstyle {\dfrac {dy}{dx}}}$ is undefined is ${\textstyle (x,y)={\color {blue}(0,0)}}$.