MATH 2210 - EXAM II REVIEW PROBLEMS
1) Show that f (x, y) = log(x2 + y 2 ) satisfies the Laplace equation
∂2f
∂2f
+
= 0.
∂x2
∂y 2
2) The curve (t, t2 , t3 ) lies on the surface z = xy. In particular, the tangent line to the curve
at any point lies in the tangent plane to the surface. Verify this statement for t = 1.
Solution: A tangent vector at (t, t2 , t3 ) is (1, 2t, 3t2 ). For t = 1 we have a point P = (1, 1, 1)
on the curve and v = (1, 2, 3) is a tangent vector. Thus x = 1 + t, y = 1 + 2t and z = 1 + 3t
are parametric equations of the tangent line.
On the other hand, the tangent plane of the surface z = f (x, y) = xy at a point (x0 , y0 , z0 ),
where z0 = f (x0 , y0 ), is z − z0 = a(x − x0 ) + b(y − y0 ) where a and b are the partial derivatives
of f (x, y) at (x0 , y0 ). In our case x0 = y0 = z0 = 1, thus z = x + y − 1 is the equation of the
tangent plane. Substituting x = 1 + t, y = 1 + 2t and z = 1 + 3t into z = x + y − 1, since
1 + 3t = (1 + t) + (1 + 2t) − 1
for all t, all points on the line are contained in the plane.
p
3) Let f (x, y) = x2 − y 2 . Approximate f (1.01, −0.02).
Solution: The point here is that (1.01, −0.02) ≈ (1, 0) and f (1, 0) = 1, easy to compute.
Around (1, 0) we can approximate f (x, y) with the linear function whose graph is the tangent
plane. This function is given by z −1 = a(x−1)+b(y −0) where a and bpare partial derivatives
2
2
of f (x, y) evaluated at (1, 0). Since fx = x/z and fx = −y/z
p (z = x − y ) we see that
a = 1 and b = 0. Thus p
z = x is the linear approximation, so 1.012 − (−0.02)2 ≈ 1.01.
My calculator gives 1.012 − (−0.02)2 ≈ 1.009801961. Thus the error of the linear approximation is less than 0.0002 and this is comparable to the square of the distance between
(1, 0) and (1.01, −0.02) (=0.0005).
√
4) The directional derivative of a√function f (x, y) at the point (1, 1) is 2 in the direction
of the unit vector ( √12 , √12 ), and 5 in the direction of the unit vector ( √15 , √25 ). Find the
gradient of the function f (x, y) at the point (1, 1).
5) Find and determine the nature of all critical points of the function f (x, y) = 2x4 −x2 +3y 2 .
6) Sand is pouring onto a conical pile in such a way that at a certain instant the height is
100 inches and increasing at 3 inches per minute and the radius is 40 inches and increasing
at 2 inches per minute. How fast is the volume increasing at that instant?
1
2
MATH 2210 - EXAM II REVIEW PROBLEMS
Solution: Let r be the radius of the base of the cone, and h the height of the cone. Then the
volume of the cone is V = 31 πr2 h i.e. the volume is a function of r and h. In the situation at
hand, r and h (and therefore V ) are functions of time t. The chain rule gives
dV
∂V dh ∂V dr
1
dh 2
dr
=
·
+
·
= πr2 ·
+ πrh · .
dt
∂h dt
∂r dt
3
dt
3
dt
We are given that dh/dt = 3 and dr/dt = 2 at a certain instant when h = 100 and r = 40.
Thus, at that instant, the volume of the cone increases
dV
1
2
= π · 402 · 3 + π · 40 · 100 · 2.
dt
3
3
cubic inches per minute.