Gradient Vectors: Basics & Computation

1) Compute the gradient of h(x, y) = x² + 3y² at (2, 1).

(4, 6)

(2, 3)

(4, 3)

(2, 6)

First, find the general gradient. The partial derivative with respect to x is found by treating y as constant: ∂h/∂x = 2x. The partial derivative with respect to y is found by treating x as constant: ∂h/∂y = 6y. So, ∇h = (2x, 6y). Now, evaluate this at the point (2, 1). For the x-component: 2 * 2 = 4. For the y-component: 6 * 1 = 6. Therefore, the gradient at (2,1) is (4, 6).

Explanation

First, find the general gradient. The partial derivative with respect to x is found by treating y as constant: ∂h/∂x = 2x. The partial derivative with respect to y is found by treating x as constant: ∂h/∂y = 6y. So, ∇h = (2x, 6y). Now, evaluate this at the point (2, 1). For the x-component: 2 * 2 = 4. For the y-component: 6 * 1 = 6. Therefore, the gradient at (2,1) is (4, 6).

2) Let f(x, y) = 3x² - 4y. What is the gradient vector of f at the point (1, 2)?

<6, -4>

<3, -4>

<6, -8>

<6x, -4>

To find the gradient vector, we first need to compute the partial derivatives of the function with respect to x and y. The partial derivative with respect to x is found by treating y as a constant, which gives us 6x. The partial derivative with respect to y is found by treating x as a constant, which gives us -4. This establishes the gradient function as <6x, -4>. Evaluating at (1,2) gives (6*1, -4) = (6, -4).

Explanation

To find the gradient vector, we first need to compute the partial derivatives of the function with respect to x and y. The partial derivative with respect to x is found by treating y as a constant, which gives us 6x. The partial derivative with respect to y is found by treating x as a constant, which gives us -4. This establishes the gradient function as <6x, -4>. Evaluating at (1,2) gives (6*1, -4) = (6, -4).

3) Find the directional derivative of the function g(x, y) = x² + y² at the point (1, 2) in the direction of the vector v = (3, 4).

22/5

11/5

22

11

The directional derivative is found using the formula D_u f = ∇f · u, where u is the unit vector in the direction of v. First, compute the gradient of g: ∇g = (2x, 2y). At the point (1,2), ∇g = (2*1, 2*2) = (2, 4). Next, find the unit vector u in the direction of v = (3,4). The magnitude of v is √(3² + 4²) = √(9+16) = √25 = 5. The unit vector u is v divided by its magnitude: u = (3/5, 4/5). Now, compute the dot product: ∇g · u = (2)*(3/5) + (4)*(4/5) = 6/5 + 16/5 = 22/5.

Explanation

The directional derivative is found using the formula D_u f = ∇f · u, where u is the unit vector in the direction of v. First, compute the gradient of g: ∇g = (2x, 2y). At the point (1,2), ∇g = (2*1, 2*2) = (2, 4). Next, find the unit vector u in the direction of v = (3,4). The magnitude of v is √(3² + 4²) = √(9+16) = √25 = 5. The unit vector u is v divided by its magnitude: u = (3/5, 4/5). Now, compute the dot product: ∇g · u = (2)*(3/5) + (4)*(4/5) = 6/5 + 16/5 = 22/5.

4) What is the directional derivative of f(x, y) = 3x - 2y at the point (5, 1) in the direction of the vector (1, 1)?

1/√2

√2

1

2

First, compute the gradient of f. The partial derivative with respect to x is 3, and with respect to y is -2. So, ∇f = (3, -2) everywhere, including at (5,1). Next, we need the unit vector in the direction of (1,1). The magnitude of (1,1) is √(1²+1²)=√2. The unit vector u is (1/√2, 1/√2). The directional derivative is the dot product: ∇f · u = (3)*(1/√2) + (-2)*(1/√2) = 3/√2 - 2/√2 = 1/√2.

Explanation

First, compute the gradient of f. The partial derivative with respect to x is 3, and with respect to y is -2. So, ∇f = (3, -2) everywhere, including at (5,1). Next, we need the unit vector in the direction of (1,1). The magnitude of (1,1) is √(1²+1²)=√2. The unit vector u is (1/√2, 1/√2). The directional derivative is the dot product: ∇f · u = (3)*(1/√2) + (-2)*(1/√2) = 3/√2 - 2/√2 = 1/√2.

5) Geometrically, the gradient vector of a function at a point points in which direction?

Steepest decrease

Steepest increase

No change

Tangent to level curve

By definition, the gradient vector of a function at a given point points in the direction where the function increases most rapidly. The magnitude of the gradient gives the rate of increase in that direction. The opposite direction (negative gradient) points in the direction of steepest decrease.

Explanation

By definition, the gradient vector of a function at a given point points in the direction where the function increases most rapidly. The magnitude of the gradient gives the rate of increase in that direction. The opposite direction (negative gradient) points in the direction of steepest decrease.

6) If the gradient of a function at a point is the zero vector, what can be concluded?

The function has a local maximum at that point.

The function has a local minimum at that point.

The function is constant in every direction from that point.

The point is a critical point, but could be a max, min, or saddle.

A point where the gradient vector is zero is called a critical point. At such a point, the function could have a local maximum, a local minimum, or a saddle point. Further analysis (using the second derivative test, for example) is needed to determine which one it is.

Explanation

A point where the gradient vector is zero is called a critical point. At such a point, the function could have a local maximum, a local minimum, or a saddle point. Further analysis (using the second derivative test, for example) is needed to determine which one it is.

7) The directional derivative of a function f in the direction of a unit vector u can be expressed as:

The magnitude of the gradient times the cosine of the angle between ∇f and u.

The magnitude of the gradient times the sine of the angle between ∇f and u.

The dot product of the gradient and u.

Both A and C

The directional derivative is defined as D_u f = ∇f · u. Using the geometric definition of the dot product, this equals the magnitude of ∇f times the magnitude of u times the cosine of the angle θ between them. Since u is a unit vector, its magnitude is 1, so D_u f = ||∇f|| cos θ. Therefore, both statements A and C are correct expressions for the directional derivative.

Explanation

The directional derivative is defined as D_u f = ∇f · u. Using the geometric definition of the dot product, this equals the magnitude of ∇f times the magnitude of u times the cosine of the angle θ between them. Since u is a unit vector, its magnitude is 1, so D_u f = ||∇f|| cos θ. Therefore, both statements A and C are correct expressions for the directional derivative.

8) For the function f(x, y) = x² + 2y², at the point (1, 1), in which direction does the function increase most rapidly?

(1,2)

(2,4)

(1,1)

(2,2)

The direction of steepest increase is given by the gradient vector. Compute the gradient: ∇f = (2x, 4y). At the point (1,1), ∇f = (2*1, 4*1) = (2, 4). This vector points in the direction of steepest increase. Note that we do not need to normalize it to find the direction; the gradient itself gives the direction.

Explanation

The direction of steepest increase is given by the gradient vector. Compute the gradient: ∇f = (2x, 4y). At the point (1,1), ∇f = (2*1, 4*1) = (2, 4). This vector points in the direction of steepest increase. Note that we do not need to normalize it to find the direction; the gradient itself gives the direction.

9) The gradient of f at point P is <4, -3>. What is the directional derivative of f at P in the direction of the vector v = <1, 0>?

4

-3

5

1

We apply the definition of the directional derivative using the gradient. The direction vector v = <1, 0> is already a unit vector because its magnitude is √(1² + 0²) = 1. The directional derivative is the dot product of the gradient <4, -3> and the direction vector <1, 0>. Calculating the dot product gives (4 * 1) + (-3 * 0), which simplifies to 4 + 0 = 4. This confirms that the directional derivative in the direction of the positive x-axis is simply the partial derivative with respect to x.

Explanation

We apply the definition of the directional derivative using the gradient. The direction vector v = <1, 0> is already a unit vector because its magnitude is √(1² + 0²) = 1. The directional derivative is the dot product of the gradient <4, -3> and the direction vector <1, 0>. Calculating the dot product gives (4 * 1) + (-3 * 0), which simplifies to 4 + 0 = 4. This confirms that the directional derivative in the direction of the positive x-axis is simply the partial derivative with respect to x.

10) At a point on a level curve of a function, the gradient vector is:

Tangent to the level curve.

Perpendicular to the level curve.

Zero

Undefined

Level curves are sets where the function is constant. The gradient points in the direction of greatest increase, which is perpendicular to the level curve (the direction where the function does not change is along the level curve). Therefore, the gradient is perpendicular (normal) to the level curve at that point.

Explanation

Level curves are sets where the function is constant. The gradient points in the direction of greatest increase, which is perpendicular to the level curve (the direction where the function does not change is along the level curve). Therefore, the gradient is perpendicular (normal) to the level curve at that point.

11) If the directional derivative of a function is zero in a particular direction, what does that indicate about the gradient and that direction?

The gradient is zero.

The gradient is perpendicular to that direction.

The gradient is parallel to that direction.

The function is constant in that direction.

The directional derivative D_u f = ∇f · u = ||∇f|| cos θ, where θ is the angle between ∇f and u. If this is zero, then either ||∇f|| = 0 (gradient is zero) or cos θ = 0, meaning θ = 90 degrees. So, the gradient is either zero or perpendicular to the direction u. Since the zero vector is technically perpendicular to every direction, we can say the gradient is perpendicular to that direction.

Explanation

The directional derivative D_u f = ∇f · u = ||∇f|| cos θ, where θ is the angle between ∇f and u. If this is zero, then either ||∇f|| = 0 (gradient is zero) or cos θ = 0, meaning θ = 90 degrees. So, the gradient is either zero or perpendicular to the direction u. Since the zero vector is technically perpendicular to every direction, we can say the gradient is perpendicular to that direction.

12) For a function of three variables, f(x, y, z) = xyz, what is the gradient at the point (1, 2, 3)?

(6,3,2)

(3,2,1)

(2,3,6)

(1,2,3)

The gradient for three variables is (∂f/∂x, ∂f/∂y, ∂f/∂z). Compute partial derivatives: ∂f/∂x = yz, ∂f/∂y = xz, ∂f/∂z = xy. At (1,2,3): ∂f/∂x = 2*3 = 6; ∂f/∂y = 1*3 = 3; ∂f/∂z = 1*2 = 2. So, ∇f = (6, 3, 2).

Explanation

The gradient for three variables is (∂f/∂x, ∂f/∂y, ∂f/∂z). Compute partial derivatives: ∂f/∂x = yz, ∂f/∂y = xz, ∂f/∂z = xy. At (1,2,3): ∂f/∂x = 2*3 = 6; ∂f/∂y = 1*3 = 3; ∂f/∂z = 1*2 = 2. So, ∇f = (6, 3, 2).

13) If the gradient of f at a point (a, b) is <3, -4>, in which direction does f increase most rapidly?

<3,-4>

<-3,4>

<4,3>

<0.6,-0.8>

We apply the significance of the gradient vector regarding rates of change. The theorem states that the maximum value of the directional derivative occurs when the direction vector is in the same direction as the gradient vector. Consequently, the function f increases most rapidly in the direction of the gradient itself. Since the gradient is given as <3, -4>, this vector represents the direction of greatest increase. While option D is the unit vector in the same direction, the gradient vector itself <3, -4> correctly identifies the direction.

Explanation

We apply the significance of the gradient vector regarding rates of change. The theorem states that the maximum value of the directional derivative occurs when the direction vector is in the same direction as the gradient vector. Consequently, the function f increases most rapidly in the direction of the gradient itself. Since the gradient is given as <3, -4>, this vector represents the direction of greatest increase. While option D is the unit vector in the same direction, the gradient vector itself <3, -4> correctly identifies the direction.

14) For the function f(x, y) = 3x² + 2y³, what is the gradient ∇f?

(6x, 6y²)

(3x, 2y²)

(6x, 2y²)

(3x², 2y³)

To find the gradient of the function f(x, y) = 3x² + 2y³, we need to compute its partial derivatives. First, we find the partial derivative with respect to x, which is done by treating y as a constant and differentiating 3x². The derivative of 3x² with respect to x is 6x. Next, we find the partial derivative with respect to y, which is done by treating x as a constant and differentiating 2y³. The derivative of 2y³ with respect to y is 6y². Therefore, the gradient vector ∇f is the vector of these partial derivatives: (6x, 6y²).

Explanation

To find the gradient of the function f(x, y) = 3x² + 2y³, we need to compute its partial derivatives. First, we find the partial derivative with respect to x, which is done by treating y as a constant and differentiating 3x². The derivative of 3x² with respect to x is 6x. Next, we find the partial derivative with respect to y, which is done by treating x as a constant and differentiating 2y³. The derivative of 2y³ with respect to y is 6y². Therefore, the gradient vector ∇f is the vector of these partial derivatives: (6x, 6y²).

15) For the function f(x, y) = 4x + 7y, what is the gradient ∇f?

(4, 7)

(4x, 7y)

(4, 0)

(0, 7)

The gradient is found by computing the partial derivatives. The partial derivative of f with respect to x treats y as a constant. The derivative of 4x is 4, and the derivative of 7y with respect to x is 0. So, ∂f/∂x = 4. The partial derivative with respect to y treats x as a constant. The derivative of 4x with respect to y is 0, and the derivative of 7y is 7. So, ∂f/∂y = 7. Therefore, the gradient vector is (4, 7).

Explanation

The gradient is found by computing the partial derivatives. The partial derivative of f with respect to x treats y as a constant. The derivative of 4x is 4, and the derivative of 7y with respect to x is 0. So, ∂f/∂x = 4. The partial derivative with respect to y treats x as a constant. The derivative of 4x with respect to y is 0, and the derivative of 7y is 7. So, ∂f/∂y = 7. Therefore, the gradient vector is (4, 7).