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

Let Find a point P(a,b,c) on the graph of such that the tangent plane to the graph of at the point P is parallel to the plane . 
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. 
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Solution 

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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 . 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: Hence, to determine points where the normal vectors are parallel, we need to solve the following two equations for (x,y): 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: 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)). 