Science:Math Exam Resources/Courses/MATH200/April 2012/Question 02 (b)

MATH200 April 2012

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  •  April 2012  •  December 2011  •  

Question 02 (b)
Let $\displaystyle {}z=f(x,y)=\ln(4x^{2}+y^{2})$ Find a point P(a,b,c) on the graph of $\displaystyle z=f(x,y)$ such that the tangent plane to the graph of $\displaystyle z=f(x,y)$ at the point P is parallel to the plane $\displaystyle 2x+2y-z=3$.

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
Consider that if the tangent plane is parallel to another plane Q, then the normal vectors between the tangent plane and Q must also be parallel.

Hint 2
The normal vector to the plane ax + by + cz = d is [a,b,c]. The normal vector to a function z = f(x,y) is given by the gradient of the function F(x,y,z) = f(x,y) - z.

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. Two planes are parallel if their normal vectors are parallel. Hence, we will evaluate the normal vectors for the given plane and the function ƒ to determine the points where they are parallel. By inspection of the equation of the plane 2x + 2y - z = 3, we can see that is normal vector is $\mathbf {n} _{1}=[2,2,-1]^{T}$. The normal vector to the function ƒ at the point (x,y) is given by computing the gradient to F(x,y,z) = ƒ(x,y) - z: $\nabla F(x,y,z)=\left[F_{x}(x,y,z),F_{y}(x,y,z),F_{z}(x,y,z)\right]^{T}=\left[{\frac {8x}{4x^{2}+y^{2}}},{\frac {2y}{4x^{2}+y^{2}}},-1\right]^{T}.$ Hence, to determine points where the normal vectors are parallel, we need to solve the following two equations for (x,y): ${\frac {8x}{4x^{2}+y^{2}}}=2,\quad {\frac {2y}{4x^{2}+y^{2}}}=2$ Taking the difference of the two equations gives us the condition y = 4x. Taking this result and plugging it into the first equation above gives: ${\begin{aligned}{\frac {8x}{4x^{2}+y^{2}}}=2\quad \rightarrow \quad 8x&=8x^{2}+2y^{2}\\8x&=8x^{2}+32x^{2}\\x&=5x^{2}\end{aligned}}$ This equation has two solutions: x = 0 and x = 1/5. Notice that the x = 0 is invalid since the corresponding value of y is y = 4x = 0 and ƒ is undefined at (0,0). Therefore, the only valid solution is x = 1/5 which corresponds to y = 4x = 4/5. The corresponding value of z is z = ƒ(1/5,4/5) = ln(4/5). Therefore, the tangent plane to ƒ is parallel to 2x + 2y - z = 3 at the point P(a,b,c) = (1/5,4/5,ln(4/5)).