Science:Math Exam Resources/Courses/MATH253/December 2012/Question 02 (a)

MATH253 December 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  

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Question 02 (a)
Let $\alpha$ be the acute angle between the plane $z=4x+2y$ and the plane tangent to the surface $z=x^{2}+y^{2}$ at the point (1,2,5). Find $\cos(\alpha )$.

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
The angle(s) between two planes is(are) given by the angle(s) between their respective normal vectors. Recall that there are two normal vectors for every plane.

Hint 2
Remember that the dot product between two vectors u and v satisfies $\mathbf {u} \cdot \mathbf {v} =\Vert \mathbf {u} \Vert \Vert \mathbf {v} \Vert \cos(\alpha )$ where $\Vert \mathbf {u} \Vert$ denotes the length of vector u and $\alpha$ is the angle between u and v.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. The angle between two planes is equal to the angle between their normal vectors. Using the hint, we remember that the angle $\alpha$ between two vectors u, v satisfies $\mathbf {u} \cdot \mathbf {v} =\Vert \mathbf {u} \Vert \Vert \mathbf {v} \Vert \cos(\alpha )\quad \rightarrow \quad \cos(\alpha )={\frac {\mathbf {u} \cdot \mathbf {v} }{\Vert \mathbf {u} \Vert \Vert \mathbf {v} \Vert }}$ where $\Vert \mathbf {u} \Vert$ denotes the length of vector u. The normal vector to the plane $4x+2y=z$ is (-4,-2,1). To determine the vector normal to the surface $z=x^{2}+y^{2}$ at (1,2,5), we compute the gradient of the function $f(x,y,z)=x^{2}+y^{2}-z$ and evaluate it at the point (1,2,5), $\nabla f(x,y,z)=(2x,2y,-1)\quad \rightarrow \quad \nabla f(1,2,5)=(2,4,-1)$ Using the formula above, we compute $\cos(\alpha )$, $\cos(\alpha )={\frac {(2,4,-1)\cdot (-4,-2,1)}{{\sqrt {4+16+1}}{\sqrt {16+4+1}}}}={\frac {-17}{21}}$ Notice that the sign of $\cos(\alpha )$ is negative, which implies an obtuse angle between the vectors. This may seem like a problem, but recall that every plane has two normal vectors which point in opposite directions. If we take the negative of either $\nabla f(1,2,5)$ or the normal vector to the plane $4x+2y=z$ and use it in our calculation for $\cos(\alpha )$, we get the positive result: $\color {blue}{\cos(\alpha )={\frac {17}{21}}}.$