Science:Math Exam Resources/Courses/MATH200/December 2013/Question 01 (e)

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MATH200 December 2013

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  

   Other MATH200 Exams

  •  April 2012  •  December 2011  •  

Question 01 (e)
Suppose it is known that the direction of the fastest increase of the function $\displaystyle f(x,y)$ at the origin is given by the vector $\displaystyle \langle 1,2\rangle$. Find a unit vector $\displaystyle {\textbf {u}}$ that is tangent to the level curve of $\displaystyle f(x,y)$ that passes through the origin.

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

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Solution

The direction of the fastest increase of a function is in the direction of the gradient of the function. The gradient of the function is orthogonal to the level curves of the function, therefore, we want to look for a vector that whose dot product with $<1,2>$ is $0$. Such a vector is $\pm <2,-1>$ Converting this into a unit vector, we get ${\overrightarrow {u}}=\color {blue}\pm <{\frac {2}{\sqrt {5}}},{\frac {-1}{\sqrt {5}}}>$

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