Find the gradient and directional derivative of f at P in the direction of v:
f(x,y)=ln(x^2 + y^2 +1)+ e^(2xy)
P(0,2)
v=5i-12j
Sorry about the boldface.
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)i
+ 2y/(x^2+y^2+1) + 2xe^(2xy)j
The directional derivative along v</v> is ∇f•v
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•v = <4,4/5>•<5,-12> = <20,-48/5>
Well, well, well! We have a mathematical question for Clown Bot! Don't worry, I won't let you down. Let's get down to business, problem-solver!
To find the gradient and directional derivative of f at P(0,2) in the direction of v(5i - 12j), we need to follow a few steps.
Step 1: Compute the gradient of f(x, y).
The gradient of f(x, y) is given by the partial derivatives of f with respect to x and y. Let's calculate those and see what we get:
∂f/∂x = (2x)/(x^2 + y^2 + 1) + 2ye^(2xy)
∂f/∂y = (2y)/(x^2 + y^2 + 1) + 2xe^(2xy)
Step 2: Evaluate the gradient at point P(0, 2).
Let's substitute x = 0 and y = 2 into the partial derivatives to obtain the partial derivatives at point P:
∂f/∂x = (2*0)/(0^2 + 2^2 + 1) + 2*2*e^(2*0*2) = 0 + 4 = 4
∂f/∂y = (2*2)/(0^2 + 2^2 + 1) + 2*0*e^(2*0*2) = 4/5
So, at point P(0,2), the gradient of f is ∇f = 4i + (4/5)j.
Step 3: Calculate the directional derivative in the direction of v.
The directional derivative of f in the direction of v is given by the dot product of the gradient of f and the unit vector in the direction of v, which is v/|v|. Let's calculate:
|v| = sqrt(5^2 + (-12)^2) = sqrt(169) = 13
Unit vector in the direction of v (let's call it u_v) = (5/13)i + (-12/13)j
Now, the directional derivative is given by:
D_vf(P) = ∇f(P) · u_v = (4i + (4/5)j) · ((5/13)i + (-12/13)j)
= (4 * 5/13) + (4/5 * -12/13)
= 20/13 - 48/65
= (260 - 48)/65
= 212/65
So, the gradient of f at P(0,2) in the direction of v(5i - 12j) is 212/65.
Well, buddy, I hope that put a smile on your face! If you have any more questions, feel free to juggle them my way!
To find the gradient and directional derivative of function f at point P in the direction of vector v, we can follow these steps:
1. Compute the partial derivatives of f with respect to x and y.
The partial derivative of f with respect to x is calculated as follows:
∂f/∂x = (∂/∂x)(ln(x^2 + y^2 + 1) + e^(2xy))
To find this derivative, treat y as a constant and differentiate each term separately. The derivative of ln(x^2 + y^2 + 1) with respect to x is 2x / (x^2 + y^2 + 1), and the derivative of e^(2xy) with respect to x is 2ye^(2xy).
2. Evaluate the partial derivatives at point P(0,2).
Substitute the values x=0 and y=2 in the expressions obtained in step 1 to find the values of the partial derivatives at point P.
3. Find the gradient at point P.
The gradient is a vector that contains the partial derivatives evaluated at point P. So, the gradient vector is:
∇f = (∂f/∂x, ∂f/∂y).
4. Normalize the vector v.
Normalize the vector v by dividing it by its magnitude. The magnitude of v is calculated as follows:
||v|| = √(5^2 + (-12)^2)
Divide each component of v by its magnitude to normalize it.
5. Compute the dot product of the normalized vector v and the gradient vector.
To find the directional derivative, compute the dot product of the normalized v vector (v_norm) and the gradient (∇f) at point P.
By following these steps, you will be able to calculate the gradient and directional derivative of function f at point P in the direction of vector v.
The gradient ∇f
= 2x/(x^2+y^2+1) + 2ye^(2xy)i
+ 2y/(x^2+y^2+1) + 2xe^(2xy)j
The directional derivative along v</v> is ∇f•v</v>
So, at (0,2)
Fx = 4
Fy = 4/5
and
∇f•v</v> = <4,4/5>•<5,-12> = <20,-48/5>