Gradient Vectors: Basics & Computation

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).

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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.

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).

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).