Test 1 sol.pdf

THE UNIVERSITY OF HONG KONG

DEPARTMENT OF MATHEMATICS

MATH2211 Multivariable Calculus

Test 1 Solutions

1. Using polar coordinates, we have

2. (a) The vectors (1, 4, −2) − (1, 1, 0) = (0, 3, −2) and (2, 1, 2) − (1, 1, 0) = (1, 0, 2) lie on Π . So (0, 3, −2) × (1, 0, 2) = (6, −2, −3)

is a normal vector to Π . Thus, the equation of Π has the form 6x − 2y − 3z = D. Putting x = y = 1 and z = 0, we get D = 4. This means the equation of Π is 6x − 2y − 3z = 4.

(b) From (a), (6, −2, −3) is a normal vector to Π . A unit normal vector is

3. The Cartesian coordinates of P and Q are (pj sin ϕj cos θj , pj sin ϕj sin θj , pj cos ϕj ) for j = 1, 2. The position vectors of P and Q are perpendicular if and only if their dot product is 0. Note that

(p1 sin ϕ1 cos θ1 , p1 sin ϕ1 sin θ1 , p1 cos ϕ1 ) . (p2 sin ϕ2 cos θ2 , p2 sin ϕ2 sin θ2 , p2 cos ϕ2 ) = 0 ⇔ p1 p2 sin ϕ1 sin ϕ2 cos θ1 cos θ2 + p1 p2 sin ϕ1 sin ϕ2 sin θ1 sin θ2 + p1 p2 cos ϕ1 cos ϕ2 = 0 ⇔ p1 p2 [sin ϕ1 sin ϕ2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) + cos ϕ1 cos ϕ2 ] = 0 ⇔ p1 p2 [sin ϕ1 sin ϕ2 cos (θ1 − θ2 ) + cos ϕ1 cos ϕ2 ] = 0.

From the condition tan ϕ1 tan ϕ2 cos (θ1 − θ2 ) = −1, we have

sin ϕ1 sin ϕ2 cos (θ1 − θ2 ) = − cos ϕ1 cos ϕ2 .

Hence, the expression inside the bracket of the above equation is 0. This proves the vectors are perpendicular.

4. (a) False. Consider u = v = (1, 0) in R2 . Then (0, 0) lies on both straight lines since

(0, 0) = (1, 0) − (1, 0) = u + (−1)v,

(0, 0) = 2(1, 0) − 2(1, 0) = 2u + (−2)v.

(b) True. Let f = (f1 , f2 , f3 ) and g = (g1 , g2 , g3 ). Then

h(v) = (f1 (v), f2 (v), f3 (v)) . (g1 (v), g2 (v), g3 (v))

= f1 (v)g1 (v) + f2 (v)g2 (v) + f3 (v)g3 (v).

Since f and g are continuous, their component functions fi , gj are continuous. As h is a sum of product of continuous functions, it is continuous.

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This study source was downloaded by 100000898231818 from CourseHero.com on 04-15-2025 20:47:44 GMT -05:00

5. By definition, we have

Since = ±1 is bounded and sin h tends to 0 when h tends to 0, this limit is equal to 0.

By symmetry, we also have fy (0, 0) = 0. To check whether f is diferentiable at (0, 0), we consider

= 0.

This shows f is diferentiable at (0, 0).

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This study source was downloaded by 100000898231818 from CourseHero.com on 04-15-2025 20:47:44 GMT -05:00

 

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