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

MATH200 December 2011

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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 $F(x,y,z)=f(x,y)-z=0$: $\mathbf {n} =\nabla F(x,y,z)={\begin{bmatrix}f_{x}(x,y)\\f_{y}(x,y)\\-1\end{bmatrix}}$.

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Solution 1

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 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.

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