Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (g)

MATH100 December 2010

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Question 01 (g)

Short-Answer Questions. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. Find the slope of the tangent line to the curve $\displaystyle x^{4}-x^{2}y+y^{4}=1$ 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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
You can't explicitly solve for y and write as a function of x. So what can you do?

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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. First, we must find the derivative of the curve using implicit differentiation: $\displaystyle {\begin{aligned}x^{4}-x^{2}y+y^{4}&=1\\(x^{4}-x^{2}y+y^{4})'&=(1)'\\4x^{3}-(x^{2}y)'+4y^{3}y'&=0\\4x^{3}-x^{2}y'-2xy+4y^{3}y'&=0\\y'(-x^{2}+4y^{3})&=2xy-4x^{3}\\y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\end{aligned}}$ This gives us the equation for the slope. The question is asking us for the slope at the point (-1,1) so we plug in x=-1 and y=1: $\displaystyle {\begin{aligned}y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\\&={\frac {2(-1)(1)-4(-1)^{3}}{4(1)^{3}-(-1)^{2}}}\\&={\frac {-2+4}{4-1}}\\&={\frac {2}{3}}\\\end{aligned}}$ Therefore, the slope of the tangent line to the curve at (-1,1) is 2/3.

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