Gradients, Level Curves & Level Surfaces

The gradient is the vector of partial derivatives. First compute ∂f/∂x treating y constant: derivative of x²y is 2xy, of 4y³ is 0, of -3x is -3, so ∂f/∂x = 2xy - 3. Then compute ∂f/∂y treating x constant: derivative of x²y is x², of 4y³ is 12y², of -3x is 0, so ∂f/∂y = x² + 12y². Thus ∇f(x,y) = (2xy - 3, x² + 12y²).

The gradient vector ∇f points in the direction of the greatest increase of the function. Consequently, the opposite direction, represented by the negative of the gradient vector (-∇f), points in the direction of the greatest decrease (steepest descent).

To compute the gradient ∇f = (∂f/∂x, ∂f/∂y), we need to find both partial derivatives using the chain rule. For ∂f/∂x, we treat y as a constant and differentiate sin(xy) with respect to x. The derivative of sin(u) is cos(u) times the derivative of u with respect to x. Here, u = xy, so du/dx = y. Therefore, ∂f/∂x = cos(xy) * y = y cos(xy). For ∂f/∂y, we treat x as a constant and differentiate sin(xy) with respect to y. Again using the chain rule, the derivative of sin(xy) with respect to y is cos(xy) times the derivative of xy with respect to y, which is x. So ∂f/∂y = cos(xy) * x = x cos(xy). The gradient is therefore ∇f = (y cos(xy), x cos(xy)).

First, we find the general gradient of f(x,y) = x² + y². The gradient is ∇f = (∂f/∂x, ∂f/∂y). Computing the partial derivatives, we get ∂f/∂x = 2x and ∂f/∂y = 2y, so ∇f = (2x, 2y). To find the gradient at the specific point (2,1), we substitute x = 2 and y = 1 into the gradient vector. Substituting gives us ∇f(2,1) = (2*2, 2*1) = (4, 2). Therefore, the gradient at point (2,1) is the vector (4, 2).

The gradient is ∇f = (∂f/∂x, ∂f/∂y). To find ∂f/∂x, we treat y as a constant and differentiate e^{xy} with respect to x. The derivative of e^u is e^u times du/dx. Here, u = xy, so du/dx = y. Therefore, ∂f/∂x = e^{xy} * y = y e^{xy}. Similarly, for ∂f/∂y, we treat x as a constant and differentiate e^{xy} with respect to y. The derivative is e^{xy} times the derivative of xy with respect to y, which is x, so ∂f/∂y = e^{xy} * x = x e^{xy}. Thus ∇f = (y e^{xy}, x e^{xy}).

The directional derivative Dᵥ f at a point is given by the dot product of the gradient ∇f and the unit vector in the direction of v. First, we compute the gradient: ∇f = (∂f/∂x, ∂f/∂y) = (2x, 2y). At point (1,1), the gradient is ∇f(1,1) = (2*1, 2*1) = (2, 2). Next, we find the unit vector in the direction of v = (1,0). The magnitude of v is ||v|| = sqrt(1**2 + 0**2) = 1, so the unit vector u is already (1,0). The directional derivative is D_u f = ∇f · u = (2, 2) · (1, 0) = 2*1 + 2*0 = 2. Therefore, the directional derivative is 2.

The directional derivative is computed as D_u f = ∇f · u. First, we find the gradient: ∇f = (∂f/∂x, ∂f/∂y) = (y, x). At point (2,1), the gradient is ∇f(2,1) = (1, 2). The given vector u = (3/5, 4/5) is already a unit vector because (3/5)**2 + (4/5)**2 = 9/25 + 16/25 = 25/25 = 1. Now we compute the dot product: D_u f = (1, 2) · (3/5, 4/5) = 1*(3/5) + 2*(4/5) = 3/5 + 8/5 = 11/5. The directional derivative is therefore 11/5.

The directional derivative requires the gradient and a unit direction vector. First, compute the gradient: ∇f = (∂f/∂x, ∂f/∂y) = (e^x cos(y), -e^x sin(y)). At point (0,0), we have e^0 = 1, cos(0) = 1, and sin(0) = 0, so ∇f(0,0) = (1, 0). Next, we convert the direction vector (1,1) to a unit vector. The magnitude is ||(1,1)|| = (2)**0.5. The unit vector is u = (1/(2**0.5), 1/(2**0.5)). The directional derivative is D_u f = ∇f · u = (1, 0) · (1/(2**0.5), 1/(2**0.5)) = 1*(1/(2**0.5)) + 0*(1/(2**0.5)) = 1/(2**0.5). Therefore, the directional derivative is 1/√2.

First, we compute the gradient of f(x,y) = x² + y², which is ∇f = (∂f/∂x, ∂f/∂y) = (2x, 2y). At point (1,1), the gradient is ∇f(1,1) = (2*1, 2*1) = (2, 2). The magnitude of a vector (a,b) is given by sqrt(a**2 + b**2). So the magnitude of the gradient is ||∇f(1,1)|| = sqrt(2**2 + 2**2) = sqrt(4 + 4) = sqrt(8). We can simplify sqrt(8) by factoring out 4: sqrt(8) = sqrt(4*2) = sqrt(4) * sqrt(2) = 2*(2**0.5). Therefore, the magnitude is 2√2.

The direction of greatest increase is given by the unit vector in the direction of the gradient. First, we find the gradient: ∇f = (∂f/∂x, ∂f/∂y) = (2x, 2y). At point (1,1), the gradient is ∇f(1,1) = (2, 2). The direction of greatest increase is the unit vector in this direction. To find the unit vector, we divide the gradient vector by its magnitude. The magnitude is ||(2,2)|| = (8)**0.5 = 2*(2**0.5). The unit vector is u = (2/(2*(2**0.5)), 2/(2*(2**0.5))) = (1/(2**0.5), 1/(2**0.5)). Therefore, the direction of greatest increase is (1/√2, 1/√2).

The level curve is x² + y² = 5. The gradient ∇f gives a normal vector to this curve. First, compute ∇f = (2x, 2y). At point (1,2), the normal vector is ∇f(1,2) = (2, 4). We can simplify this to (1, 2) by dividing by 2. The equation of a line with normal vector (a,b) passing through point (x₀,y₀) is a(x-x₀) + b(y-y₀) = 0. Substituting a=1, b=2, x₀=1, y₀=2 gives: 1(x-1) + 2(y-2) = 0. Expanding: x - 1 + 2y - 4 = 0. Combining terms: x + 2y - 5 = 0. Rearranging: x + 2y = 5. Therefore, the tangent line equation is x + 2y = 5.

Just as the gradient of a two-variable function is perpendicular to level curves, the gradient of a three-variable function is perpendicular (normal) to the level surfaces. Therefore, ∇F serves as the normal vector to the tangent plane of the level surface at that point.

The relationship between the gradient ∇f and the directional derivative D_u f in the direction of a unit vector u is given by the formula D_u f = ∇f · u. This dot product can be expressed as ||∇f|| * ||u|| * cos(θ), where θ is the angle between ∇f and u. Since u is a unit vector, ||u|| = 1, so D_u f = ||∇f|| cos(θ). This is precisely the definition of the projection of the gradient vector onto the direction u. In other words, the directional derivative measures how much the gradient points in the specified direction.

The directional derivative is D_u f = ||∇f|| cos(θ). The maximum possible rate of change is ||∇f||. We are given that D_u f = 0.5 * ||∇f||. Setting these equal: ||∇f|| cos(θ) = 0.5 * ||∇f||. Assuming the gradient is not zero, we can divide by ||∇f|| to get cos(θ) = 0.5. The angle whose cosine is 0.5 is 60 degrees (π/3 radians).

For a function f(x,y) of two variables, the gradient ∇f = (∂f/∂x, ∂f/∂y) defines a vector field in the xy-plane. At each point (x₀,y₀), the gradient vector ∇f(x₀,y₀) points in the direction of steepest increase of the function and its magnitude gives the rate of this steepest increase. Thus it represents, geometrically, the direction and rate of steepest ascent of the surface z = f(x,y).