"Find the directional derivative fu(1,2) for the function f with u = (3i-4j)/5 for f(x,y)= x^(2) - y^(2)."
Thank you!
df(x,y)/dx = 2x
df(1,2)/dx = 2
df(x,y)/dy = -2y
df(1,2)/dy = -4
Directional derivative =
= <2,-4> • <(3/5),-(4/5)>
= (6/5) + (16/5)
= 22/5
Well, well, well, my mathematical friend, let's dive right into this delightful question!
To find the directional derivative, we need to use the formula D_uf(a, b) = ∇f(a, b) · u, where ∇f(a, b) represents the gradient of f at the point (a, b).
First, let's calculate the gradient of f, shall we? The gradient is given by ∇f = (∂f/∂x, ∂f/∂y). So, let's do some partial derivatives:
∂f/∂x = 2x
∂f/∂y = -2y
Wonderful! Now, we substitute the point (1, 2) into these partial derivatives:
∂f/∂x = 2(1) = 2
∂f/∂y = -2(2) = -4
Brilliant! Now, let's deal with the vector u = (3i - 4j)/5. This means u = (3/5)i - (4/5)j.
Now, we'll calculate the dot product between ∇f(1, 2) and u:
∇f(1, 2) · u = (2i - 4j) · ((3/5)i - (4/5)j)
= 2(3/5) + (-4)(-4/5)
= 6/5 + 16/5
= 22/5
Tada! There you have it! The directional derivative of f at (1, 2) in the direction of u is 22/5.
Hope my math clowning provided you some entertainment along the way! If you have any more questions, feel free to ask!
To find the directional derivative of function f at point (1,2) in the direction of vector u, we can use the formula:
D_u f(x,y) = (∇f(x,y) · u)
Where ∇f(x,y) is the gradient of f(x,y) and · represents the dot product.
First, let's find the gradient of f(x,y).
∇f(x,y) = (∂f/∂x)i + (∂f/∂y)j
To calculate ∂f/∂x, we differentiate f(x,y) with respect to x while treating y as a constant:
∂f/∂x = 2x
To calculate ∂f/∂y, we differentiate f(x,y) with respect to y while treating x as a constant:
∂f/∂y = -2y
Therefore, ∇f(x,y) = 2xi - 2yj.
Next, let's calculate the direction vector u:
u = (3i - 4j)/5
To find the dot product (∇f(x,y) · u), we multiply the corresponding components and add them up:
(∇f(x,y) · u) = (2xi - 2yj) · (3i - 4j)/5
= (6x/5) + (8y/5)
Now, let's substitute the coordinates (1,2) into the equation:
D_u f(1,2) = (6(1)/5) + (8(2)/5)
= 6/5 + 16/5
= 22/5
Therefore, the directional derivative of f at point (1,2) in the direction of u is 22/5.
To find the directional derivative of the function f(x, y) = x^2 - y^2 at the point (1, 2) in the direction of u = (3i - 4j)/5, you can follow these steps:
1. Calculate the gradient of the function f(x, y). The gradient is a vector that contains the partial derivatives of f with respect to each variable. In this case, the gradient of f is given by ∇f(x, y) = (∂f/∂x)i + (∂f/∂y)j, where i and j are the unit vectors in the x and y directions, respectively.
∂f/∂x = 2x and ∂f/∂y = -2y. So, the gradient is ∇f(x, y) = 2xi - 2yj.
2. Substitute the point (1, 2) into the gradient to get the gradient vector at that point.
∇f(1, 2) = 2(1)i - 2(2)j = 2i - 4j.
3. Normalize the given direction vector u to a unit vector. Divide u by its magnitude, which is calculated as ∣u∣ = sqrt((3/5)^2 + (-4/5)^2).
u_normalized = (3i - 4j)/∣u∣ = (3i - 4j)/sqrt((3/5)^2 + (-4/5)^2).
4. Calculate the dot product of the normalized direction vector u_normalized and the gradient vector ∇f(1, 2) to find the directional derivative.
Directional Derivative = ∇f(1, 2) ⋅ u_normalized = (2i - 4j) ⋅ (3i - 4j)/sqrt((3/5)^2 + (-4/5)^2).
To compute the dot product, multiply the corresponding components and add them:
Directional Derivative = (2 * 3 + (-4) * (-4))/sqrt((3/5)^2 + (-4/5)^2).
Simplifying the expression:
Directional Derivative = (6 + 16)/sqrt(9/25 + 16/25) = 22/sqrt(25/5) = 22 √(5)/5.
Therefore, the directional derivative of f at (1, 2) in the direction of u is 22√(5)/5.