The directional derivative of z = f(x,y) at (2,1) in the direction toward the point (1,3) is -2/sqrt(5), and the directional derivative in the direction toward the point (5,5) is 1. Compute dz/dx and dz/dy at (2,1).
I have no idea how to solve this problem. Help, please? Thank you.
From point (2,1) to (1,3):
u = <1-2,3-1> = <-1,2>
Note that vector u above is not a unit vector, so we need to make this a unit vector by dividing it with its magnitude.
u = <-1,2>/sqrt(5)
<dz/dx,dz/dy>•<-1/sqrt(5),2/sqrt(5)> = -2/sqrt(5)
-dz/dx + 2dz/dy = -2. ..(1)
From point (2,1) to (5,5):
v = <5-2,5-1> = <3,4>
Again, this is not a unit vector.
v = <3,4>/5 = <3/5,4/5>
<dz/dx,dz/dy>•<3/5,4/5> = 1
3dz/dx + 4dz/dy = 5. ..(2)
Using elimination and substition to both equations (1) and (2), you'll get the value for dz/dx and dz/dy
To solve this problem, we can use the concept of directional derivatives. The directional derivative of a function f(x, y) in the direction of a vector u = (a, b) is given by the dot product of the gradient of f(x, y) and the unit vector of u:
D(u) = ∇f(x, y) ⋅ (â, b̂)
Given that the directional derivative in the direction toward the point (1, 3) is -2/√5 and in the direction toward the point (5, 5) is 1, we can set up the following equations:
∇f(2, 1) ⋅ (1/√10, 3/√10) = -2/√5
∇f(2, 1) ⋅ (5/√50, 5/√50) = 1
To compute dz/dx and dz/dy at (2, 1), we need to find the partial derivatives ∂z/∂x and ∂z/∂y.
Let's solve the equations:
First, we can find the gradient of f(x, y):
∇f(x, y) = (∂f/∂x, ∂f/∂y)
Now, let's compute the dot product of the gradient with the given unit vectors:
(∂f/∂x, ∂f/∂y) ⋅ (1/√10, 3/√10) = -2/√5
(∂f/∂x, ∂f/∂y) ⋅ (5/√50, 5/√50) = 1
Simplifying further, we get two equations:
∂f/∂x + 3∂f/∂y = -2/√5 ----(1)
5∂f/∂x + 5∂f/∂y = 1 ----(2)
To solve this system of equations, we can multiply equation (1) by 5:
5∂f/∂x + 15∂f/∂y = -2√5
Now we can subtract equation (2) from this new equation:
(5∂f/∂x + 15∂f/∂y) - (5∂f/∂x + 5∂f/∂y) = -2√5 - 1
Simplifying the equation further, we get:
10∂f/∂y = -2√5 - 1
Let's solve for ∂f/∂y:
∂f/∂y = (-2√5 - 1)/10
Now, plugging this value back into equation (1):
∂f/∂x + 3((-2√5 - 1)/10) = -2/√5
Simplifying this equation, we can solve for ∂f/∂x:
∂f/∂x = -2/√5 - 3((-2√5 - 1)/10)
Finally, we have ∂f/∂x and ∂f/∂y at (2, 1):
∂f/∂x = -2/√5 - 3((-2√5 - 1)/10)
∂f/∂y = (-2√5 - 1)/10
To compute dz/dx and dz/dy at (2,1), we can use the relationship between the directional derivative and the partial derivatives. The directional derivative of z = f(x, y) at a point (a, b) in the direction of a unit vector u = (u1, u2) can be expressed as:
Df(a, b) = ∇f(a, b) · u
where ∇f(a, b) is the gradient of f(x, y) at (a, b), and the dot operator represents the dot product between the gradient vector and the unit vector.
To find dz/dx and dz/dy, we need to calculate the partial derivatives of f(x, y) with respect to x and y at (2,1).
Let's first find the unit vector in the direction toward the point (1,3). We can calculate the direction vector from (2,1) to (1,3), which is v = (1-2, 3-1) = (-1, 2). Next, we normalize this vector to obtain a unit vector u:
u = v / ||v|| = (-1, 2) / sqrt((-1)^2 + 2^2) = (-1, 2) / sqrt(1 + 4) = (-1, 2) / sqrt(5)
Given that the directional derivative in the direction toward (1,3) is -2/sqrt(5), we can now establish the relationship:
Df(2, 1) = ∇f(2, 1) · u = -2/sqrt(5)
Similarly, we can find the unit vector in the direction toward (5,5). The direction vector is v = (5-2, 5-1) = (3, 4), and the unit vector is:
u = v / ||v|| = (3, 4) / sqrt(3^2 + 4^2) = (3, 4) / 5 = (3/5, 4/5)
Given that the directional derivative in the direction toward (5,5) is 1, we have:
Df(2, 1) = ∇f(2, 1) · u = 1
Now, we have two equations:
1) ∇f(2, 1) · (-1, 2) / sqrt(5) = -2/sqrt(5)
2) ∇f(2, 1) · (3/5, 4/5) = 1
We can solve these equations simultaneously to find ∇f(2, 1), which represents the gradient vector. The gradient vector contains the partial derivatives dz/dx and dz/dy.
By simplifying the equations, we have:
-∇f(2, 1) · (1/sqrt(5), -2/sqrt(5)) = -2/sqrt(5)
∇f(2, 1) · (3/5, 4/5) = 1
Solving these equations simultaneously will yield the values of dz/dx and dz/dy at (2,1).