Suppose that the height of a hill above sea level is given by z= 1000 -0.01x2 – 0.02y2. if you are at the point (60,1000) in what direction is the elevation changing fastest? What is the maximum rate of change of the elevation at this point?
Well, with a function like that, we certainly won't be climbing any mountains for comedy material! But fear not, I'll do my best to entertain you while answering your question!
To determine the direction of the fastest elevation change, we need to find the gradient vector of the function. Let's calculate that, shall we?
The gradient vector, ∇z, represents the direction of steepest ascent at any given point. It is given by the partial derivatives of the function with respect to x and y, respectively:
∂z/∂x = -0.02x
∂z/∂y = -0.04y
At the point (60, 1000), the partial derivatives become:
∂z/∂x = -0.02(60) = -1.2
∂z/∂y = -0.04(1000) = -40
So, the gradient vector is (-1.2, -40). However, to find the direction of steepest ascent, we need to take the negative of this vector. Therefore, the direction of fastest elevation change is approximately (1.2, 40).
Now, to determine the maximum rate of change of the elevation at this point, we can use the magnitude of the gradient vector:
|∇z| = √((-1.2)^2 + (-40)^2) ≈ 40.198
So, the maximum rate of change of the elevation at this point is approximately 40.198 units.
To determine the direction in which the elevation is changing fastest and the maximum rate of change at the given point (60, 1000), we need to find the gradient vector and its magnitude.
The gradient vector, ∇z, represents the direction of the steepest ascent on the surface given by the function z(x, y), while its magnitude, ||∇z||, gives the rate of change in that direction.
First, let's find the partial derivatives:
∂z/∂x = -0.02x
∂z/∂y = -0.04y
Now we can calculate the gradient vector:
∇z = (∂z/∂x, ∂z/∂y) = (-0.02x, -0.04y)
Substituting the coordinates of the given point (60, 1000) into the gradient vector, we have:
∇z(60, 1000) = (-0.02 * 60, -0.04 * 1000) = (-1.2, -40)
So, at the point (60, 1000), the elevation is changing fastest in the direction (-1.2, -40).
To find the maximum rate of change, we calculate the magnitude of the gradient vector:
||∇z|| = √((-1.2)² + (-40)²) = √(1.44 + 1600) = √(1601.44) ≈ 40.016
Therefore, the maximum rate of change of the elevation at the point (60, 1000) is approximately 40.016 units.
To find the direction in which the elevation is changing fastest and the maximum rate of change at a given point, we need to calculate the gradient vector of the hill function at that point.
The gradient vector is a vector that points in the direction of the steepest ascent of a function, and its magnitude represents the rate of change of the function in that direction.
In this case, the hill function is given by z = 1000 - 0.01x^2 - 0.02y^2. To find the gradient vector, we need to calculate the partial derivatives with respect to x and y.
Taking the partial derivative with respect to x, we get:
∂z/∂x = -0.02x
Taking the partial derivative with respect to y, we get:
∂z/∂y = -0.04y
Now, we can evaluate the gradient vector at the point (60, 1000) by substituting the coordinates into the partial derivatives:
∂z/∂x = -0.02(60) = -1.2
∂z/∂y = -0.04(1000) = -40
So, the gradient vector at the point (60, 1000) is (-1.2, -40).
The direction of the steepest ascent is the direction of this gradient vector, which is (-1.2, -40). However, since we are interested in the direction, we can represent it as a unit vector by dividing the gradient vector by its magnitude.
The magnitude of the gradient vector is given by:
|∇z| = √((-1.2)^2 + (-40)^2) = 40.008
Dividing the gradient vector by its magnitude, we obtain the unit vector in the direction of steepest ascent:
(-1.2/40.008, -40/40.008) ≈ (-0.029999, -0.999)
So, the elevation is changing fastest in the direction of (-0.029999, -0.999).
To find the maximum rate of change at this point, we need to find the magnitude of the gradient vector. We already calculated it earlier:
|∇z| = √((-1.2)^2 + (-40)^2) = 40.008
Therefore, the maximum rate of change of the elevation at the point (60, 1000) is approximately 40.008.
z = 1000 - .01 x^2 - .02 y^2
dz = -.02 x dx -.04 y dy
dz= -.02(60)dx - .04(1000)dy
dz = -1.2 dx - 40 dy
move in y direction , about 40 times as effective as x motion
dz/dy = -40