Vector Calculus MCQs

1

Advanced Mathematics • Vector Calculus

Find the directional derivative D_u f(x,y) at (x,y) = (1,2), where f(x,y) = x^3 - 3xy + 4y^2 and u is the unit vector in the direction theta = pi/6. Find the directional derivative D_u f(x,y) at (x,y) = (1,2), where f(x,y) = x^3 - 3xy + 4y^2 and u is the unit vector in the direction theta = pi/6.

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A. (13 - 3sqrt(3))/2 (13 - 3sqrt(3))/2 B. 10 10 C. 3sqrt(3) 3sqrt(3) D. 0 0

सही उत्तर: A Correct Answer: A

At (1,2), grad f = (3x^2 - 3y, -3x + 8y) = (-3,13). Dotting it with (cos(pi/6), sin(pi/6)) gives (13 - 3sqrt(3))/2.

2

Advanced Mathematics • Vector Calculus

Find the length of the arc of the circular helix with vector equation r(t) = cos(t)i + sin(t)j + tk, from (1,0,0) to (1,0,2pi). Find the length of the arc of the circular helix with vector equation r(t) = cos(t)i + sin(t)j + tk, from (1,0,0) to (1,0,2pi).

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A. sqrt(2)pi sqrt(2)pi B. 2sqrt(2)pi 2sqrt(2)pi C. 1 1 D. pi pi

सही उत्तर: B Correct Answer: B

The magnitude of the first derivative of r(t) is sqrt(sin^2(t) + cos^2(t) + 1) = sqrt(2). Integrating the speed from 0 to 2pi gives 2sqrt(2)pi.

3

Advanced Mathematics • Vector Calculus

Evaluate the line integral over C of (x^4 dx + xy dy), where C is the triangular curve consisting of line segments from (0,0) to (1,0), from (1,0) to (1,1), and from (1,1) to (0,0). Evaluate the line integral over C of (x^4 dx + xy dy), where C is the triangular curve consisting of line segments from (0,0) to (1,0), from (1,0) to (1,1), and from (1,1) to (0,0).

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A. 0 0 B. 12 12 C. 1/6 1/6 D. 1 1

सही उत्तर: C Correct Answer: C

Evaluating the integral along the three directed line segments and adding the results gives 1/6.

4

Advanced Mathematics • Vector Calculus

For f(x,y) = x^2y - y^3, which one of the following gives the corresponding gradient vector field? For f(x,y) = x^2y - y^3, which one of the following gives the corresponding gradient vector field?

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A. x^2 i x^2 i B. 2xy i + (x^2 - 3y^2) j 2xy i + (x^2 - 3y^2) j C. x i + (x^2 - y^2) j x i + (x^2 - y^2) j D. (x^3 - y^2) j (x^3 - y^2) j

सही उत्तर: B Correct Answer: B

The partial derivatives are f_x = 2xy and f_y = x^2 - 3y^2. Hence grad f = 2xy i + (x^2 - 3y^2)j.

5

Advanced Mathematics • Vector Calculus

If f(x,y,z) = x sin(yz), find the gradient of f at the point (1,3,0). If f(x,y,z) = x sin(yz), find the gradient of f at the point (1,3,0).

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A. (1,0,3) (1,0,3) B. (1,-1,3) (1,-1,3) C. (0,0,3) (0,0,3) D. (0,0,0) (0,0,0)

सही उत्तर: C Correct Answer: C

grad f = (sin(yz), xz cos(yz), xy cos(yz)). At (1,3,0), this becomes (0,0,3).

6

Advanced Mathematics • Vector Calculus

Consider f(x,y) = xe^y. At the point (2,0), what is the maximum rate of change? Consider f(x,y) = xe^y. At the point (2,0), what is the maximum rate of change?

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A. sqrt(5) sqrt(5) B. 4 4 C. 1 1 D. 0 0

सही उत्तर: A Correct Answer: A

The maximum directional derivative is |grad f|. Here grad f=(e^y,xe^y), which at (2,0) is (1,2), of magnitude sqrt(5).

Vector Calculus MCQs

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