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Exam 2: Solutions
Problem 1. Use the chain rule to find ∂z∂u and ∂z∂v if z = x^2 y^3 +x sin y, x = u^2 , y = uv.
Solution. ∂z ∂u
∂z ∂x
∂x ∂u
∂z ∂y
∂y ∂u
= (2xy^3 + sin y)(2u) + (3x^2 y^2 + x cos y)v
= (2u^2 u^3 v^3 + sin(uv))(2u) + (3u^4 u^2 v^2 + u^2 cos(uv))v = 4 u^6 v^3 + 2u sin(uv) + 3u^6 v^3 + u^2 v cos(uv) = 7 u^6 v^3 + 2u sin(uv) + u^2 v cos(uv). ∂z ∂v
∂z ∂x
∂x ∂v
∂z ∂y
∂y ∂v
= (2xy^3 + sin y) · 0 + (3x^2 y^2 + x cos y)u
= (3u^4 u^2 v^2 + u^2 cos(uv))u = 3u^7 v^2 + u^3 cos(uv).
Problem 2. Find the directional derivative of f (x, y, z) = x^2 y^3 + √ xz at (1, − 2 , 3) in the direction of −→a = 5
j +
k.
Solution. The unit vector in the direction of −→a is
−→u =
−→a ‖−→a ‖
j +
k.
Then
D−→u f (1, − 2 , 3) = ∇f (1, − 2 , 3) · −→u
=
2 xy^3 +
z 2
x
i + 3x^2 y^2
j +
x 2
z
k
(1,− 2 ,3)
j +
k
i + 12
j +
k
j +
k
Problem 3. Find the equation of the tangent plane to the surface z = x^2 y at the point (1, 2 , 2). 1
Solution. The equation of the tangent plane to the surface z = f (x, y) at the point (x 0 , y 0 , z 0 ) is
z = z 0 + fx(x 0 , y 0 )(x − x 0 ) + fy(x 0 , y 0 )(y − y 0 ).
We have fx(x, y) = 2xy, fy(x, y) = x^2 , fx(1, 2) = 2 · 1 · 2 = 4, fy(1, 2) = 12 = 1. Thus, the tangent plane is z = 2 + 4(x − 1) + 1(y − 2), i.e.,
z = 4x + y − 4.
Problem 4. The temperature, T , at a point in space is given by T (x, y, z) = 400xyz^2 degrees Celsius. Use the Lagrange multiplier method to find the highest temperature on the sphere x^2 + y^2 + z^2 = 1.
Solution. The problem is to maximize T (x, y, z) = 400xyz^2 which is subject to the constraint g(x, y, z) = x^2 + y^2 + z^2 − 1 = 0. The Lagrange equations Tx = λgx, Ty = λgy, Tz = λgz become 400yz^2 = 2 λx, 400xz^2 = 2λy, 800xyz = 2λz. Eliminating λ from these equations, we obtain 200 yz^2 x
200 xz^2 y
= 400xy.
(Note that z 6 = 0, otherwise it follows from the Lagrange equations that x = y = z = 0 which is impossible for the given constraint.) The first of the two equations above gives x^2 = y^2 , and the second one gives z^2 = 2y^2. From the constraint equation we obtain y^2 +y^2 +2y^2 = 1, i.e., 4 y^2 = 1. Thus y = 12 or y = −^12. Correspondingly, x = 12 or x = −^12 ,
and z =
√ 2 2 or^ z^ =^ −
√ 2 2.^ The maximal value of the temperature T (x, y, z) on the sphere x^2 + y^2 + z^2 = 1 is therefore
T
= 50 (degrees Celsius).
Problem 5. Find the volume of the solid in the first octant enclosed by z = 4 − y^2 , z = 0, x = 0, y = x, and y = 2.
Solution. The projection of the solid onto the xy-plane is the (tri- angle) domain D bounded by the lines x = 0, y = x, and y = 2. It can be described by the inequalities 0 ≤ y ≤ 2, 0 ≤ x ≤ y. Therefore the
Solution. The domain of integration D is described as a type II domain by 0 ≤ y ≤ 1, 2y ≤ x ≤ 2. It can be easily seen that D is a triangle bounded by the lines y = 0, x = 2y, and x = 2. The domain D is described also as a type I domain by 0 ≤ x ≤ 2, 0 ≤ y ≤ x 2. Therefore,
∫^1
0
∫^2
2 y
cos(x^2 ) dx dy =
∫^2
0
x 2 ∫
0
cos(x^2 ) dy dx =
∫^2
0
y cos(x^2 )
x 2 y=0 dx
∫^2
0
x 2
cos(x^2 ) dx =
∫^2
0
2 x cos(x^2 ) dx
∫^2
0
cos(x^2 ) d(x^2 ) =
sin(x^2 )
∣^2
sin 4 4
