Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (b)

MATH105 April 2013

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Question 01 (b)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Find an equation for the level curve of $\displaystyle f(x,y)=3xy^{2}+2y-1$ that goes through the point $\displaystyle (1,-2)$.

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
A level curve of a function $\displaystyle f(x,y)$ is the set of points $\displaystyle x,y$ satisfying the equation $\displaystyle f(x,y)=constant$ What does it mean for the constant, when we know that a special point $\displaystyle (a_{1},a_{2})$ satisfies this equation? Can you calculate the constant?

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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. In the equation of a level curve we want to know all $\displaystyle x,y\in \mathbb {R}$, such that $\displaystyle f(x,y)$ is constant. Since the point $\displaystyle (1,-2)$ shall lie in the level curve, we search for the set satisfying $\displaystyle f(x,y)=f(1,-2).$ This means $\displaystyle 3xy^{2}+2y-1=7$ We can simplify this equation to $\displaystyle y(3xy+2)=8$

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