Solutions to the First Midterm Exam – Multivariable Calculus
Math 53, February 25, 2011. Instructor: E. Frenkel
1. Consider the curve in R2defined by the equation
r= cos(2θ).
(a) Sketch this curve.
(b) Find the area of the region enclosed by one loop of this curve.
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2Zπ/4
−π/4
cos2(2θ)dθ =1
4Zπ/4
−π/4
(1 + cos(4θ))dθ =π
8.
2. (a) Find an equation of the surface consisting of all points in R3that are equidistant
from the point (0,0,1) and the plane z= 2.
The distance from a point P= (x, y, z) to the point (0,0,1) is px2+y2+ (z−1)2,
and the distance to the plane z= 2 is z−2. Hence we obtain the equation
px2+y2+ (z−1)2=z−2,
which gives
x2+y2+ (z−1)2= (z−2)2,
and hence
z=−x2
2−y2
2+3
2.
(b) Sketch this surface. What is it called?
This is an elliptic paraboloid which goes downward along the zaxis.
3. Show that the function x50y50
x100 +y200 does not have a limit at (x, y) = (0,0).
First let’s approach (0,0) along the y-axis. Then x= 0 and y6= 0, so we have along
this path 0/y200 = 0 which has limit 0.
Now let’s approach (0,0) along the line x=y. Then we obtain x100/(x100 +x200 ) =
1/(1 + x100) which has the limit 1 as x→0.
Since the function has two diferent limits along two different lines approaching (0,0),
the limit of this function at (0,0) does not exist.
4. Consider the function f(x, y) = xcos(y) + y2ex+x.
(a) Find the differential of this function.
df = (cos(y) + y2ex+ 1)dx + (−xsin(y)+2yex)dy.
(b) Find an equation of the tangent plane to the graph of this function at the point
(0, π, π2).
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