Science:Math Exam Resources/Courses/MATH200/December 2011/Question 01 (b)

{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}

MATH200 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7  •  Q8 (a)  •  Q8 (b) i  •  Q8 (b) ii  •  Q8 (b) iii  •  

   Other MATH200 Exams

  •  April 2012  •  December 2011  •  

Question 01 (b)
Consider the function $\displaystyle f(x,y)=e^{-x^{2}+4y^{2}}$ b) Find the equation of the tangent plane to the graph $z=f(x,y)$ at the point where $(x,y)=(2,1)$.

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 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

Remember that all vectors lying in the tangent plane to $z=f(x,y)$ are perpendicular to the normal vector to the surface at the same point. I.e., let $(x,y,z)=(x_{0},y_{0},z_{0})$ be a point on the surface of $z=f(x,y)$. If $\mathbf {n}$ is the normal vector to $z=f(x,y)$ at this point, then $\mathbf {n} \cdot \mathbf {v} =\mathbf {0} ,\quad {\mbox{where}}\quad \mathbf {v} ={\begin{bmatrix}x-x_{0}\\y-y_{0}\\z-z_{0}\end{bmatrix}}$ gives the equation of the tangent plane to the surface at the point $(x_{0},y_{0},z_{0})$.

Hint 2
In this case, the normal vector, $\mathbf {n}$, is determined by the gradient of function $F(x,y,z)=f(x,y)-z=0$: $\mathbf {n} =\nabla F(x,y,z)={\begin{bmatrix}f_{x}(x,y)\\f_{y}(x,y)\\-1\end{bmatrix}}$.

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 1

The equation of the tangent plane is given by evaluating $\mathbf {n} \cdot \mathbf {v} =0$ where $\mathbf {v} =[x-2,y-1,z-f(2,1)]^{T}$ and $\mathbf {n}$ is a normal vector to the surface $z=f(x,y)$ at $(x,y)=(2,1)$. The normal vector at the point $(x,y)$ is given by $\mathbf {n} (x,y)={\begin{bmatrix}f_{x}(x,y)\\f_{y}(x,y)\\-1\end{bmatrix}}={\begin{bmatrix}-2xe^{-x^{2}+4y^{2}}\\8ye^{-x^{2}+4y^{2}}\\-1\end{bmatrix}}\quad \rightarrow \quad \mathbf {n} (2,1)={\begin{bmatrix}-4\\8\\-1\end{bmatrix}}.$ Evaluating $\mathbf {n} \cdot \mathbf {v} =0$ gives $\mathbf {n} \cdot \mathbf {v} =-4(x-2)+8(y-1)-1(z-1)=0.$ We simplify the above equation (although it is not necessary) and get the tangent plane to $f(x,y)$ at $(2,1)$: ${\color {blue}-4x+8y-z=-1.}$

Solution 2

We can also get the tangent plane by using a linear Taylor approximation. Recall for a function $\displaystyle {}f(x,y)$ we have the linear approximation at a point $\displaystyle {}(x_{0},y_{0})$ is $f(x,y)\approx {}f(x_{0},y_{0})+{\frac {\partial {}f}{\partial {}x}}(x_{0},y_{0})(x-x_{0})+{\frac {\partial {}f}{\partial {}y}}(x_{0},y_{0})(y-y_{0}).$ We are given that we want to approximate $\displaystyle {}f(2,1)$ and so $\displaystyle {}x_{0}=2$ and $\displaystyle {}y_{0}=1$. We have ${\begin{aligned}{\frac {\partial {}f}{\partial {}x}}&=-2x{\textrm {e}}^{-x^{2}+4y^{2}}\\{\frac {\partial {}f}{\partial {}y}}&=8y{\textrm {e}}^{-x^{2}+4y^{2}}\end{aligned}}$ and so ${\begin{aligned}f(2,1)&=1\\{\frac {\partial {}f}{\partial {}x}}(2,1)&=-4\\{\frac {\partial {}f}{\partial {}y}}(2,1)&=8.\end{aligned}}$ Since $\displaystyle {}z=f(x,y)$ we have ${\begin{aligned}z&=1-4(x-2)+8(y-1)\\0&=-4(x-2)+8(y-1)-(z-1)\end{aligned}}$ as the tangent plane approximation at the desired point.

{{#incat:MER CT flag||

Click here for similar questions

MER QGH flag, MER QGQ flag, MER RS flag, MER RT flag, MER Tag Tangent plane, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

}}