Below is a contour map for a function f(x, y) that describes the topography of a park. (a) Estimate the values of f. f and fy at the point (x, y) = (0.6.0.4). (b) What is the equation for the tangent plane to the graph of f at the point corresponding to (x,y) (0.6,0.4)? (c) Estimate far, fry, fyz and fyy at the point (x, y) = (0.6.0.4).
The Correct Answer and Explanation is :
To address the problem involving the contour map of a function ( f(x, y) ) representing the topography of a park, we will estimate the values of ( f ), ( f_x ), and ( f_y ) at the point ( (x, y) = (0.6, 0.4) ), determine the equation of the tangent plane at this point, and estimate the second-order partial derivatives ( f_{xx} ), ( f_{xy} ), ( f_{yx} ), and ( f_{yy} ).
(a) Estimating ( f ), ( f_x ), and ( f_y ) at ( (0.6, 0.4) ):
- Estimating ( f(0.6, 0.4) ):
- From the contour map, locate the contour lines near ( (0.6, 0.4) ). The contour values at ( y = 0.4 ) are approximately 30 at ( x = 0.6 ), 15 at ( x = 0.4 ), and 59 at ( x = 0.8 ). Interpolating these values suggests that ( f(0.6, 0.4) \approx 30 ).
- Estimating ( f_x(0.6, 0.4) ):
- The partial derivative ( f_x ) represents the rate of change of ( f ) with respect to ( x ) at a fixed ( y ). To estimate ( f_x ) at ( (0.6, 0.4) ), consider the change in ( f ) between ( x = 0.4 ) and ( x = 0.8 ) at ( y = 0.4 ):
[
f_x(0.6, 0.4) \approx \frac{f(0.8, 0.4) – f(0.4, 0.4)}{0.8 – 0.4} = \frac{59 – 15}{0.4} = \frac{44}{0.4} = 110
]
Thus, ( f_x(0.6, 0.4) \approx 110 ).
- Estimating ( f_y(0.6, 0.4) ):
- The partial derivative ( f_y ) represents the rate of change of ( f ) with respect to ( y ) at a fixed ( x ). To estimate ( f_y ) at ( (0.6, 0.4) ), consider the change in ( f ) between ( y = 0.2 ) and ( y = 0.6 ) at ( x = 0.6 ):
[
f_y(0.6, 0.4) \approx \frac{f(0.6, 0.6) – f(0.6, 0.2)}{0.6 – 0.2} = \frac{86 – (-13)}{0.4} = \frac{99}{0.4} = 247.5
]
Thus, ( f_y(0.6, 0.4) \approx 247.5 ).
(b) Equation of the Tangent Plane at ( (0.6, 0.4) ):
The equation of the tangent plane to the surface ( z = f(x, y) ) at the point ( (x_0, y_0, z_0) ) is given by:
[
z – z_0 = f_x(x_0, y_0)(x – x_0) + f_y(x_0, y_0)(y – y_0)
]
Substituting the estimated values:
[
z – 30 = 110(x – 0.6) + 247.5(y – 0.4)
]
Simplifying:
[
z = 30 + 110(x – 0.6) + 247.5(y – 0.4)
]
This is the equation of the tangent plane at ( (0.6, 0.4) ).
(c) Estimating ( f_{xx} ), ( f_{xy} ), ( f_{yx} ), and ( f_{yy} ) at ( (0.6, 0.4) ):
- Estimating ( f_{xx}(0.6, 0.4) ):
- The second partial derivative ( f_{xx} ) represents the rate of change of ( f_x ) with respect to ( x ). To estimate ( f_{xx} ) at ( (0.6, 0.4) ), consider the change in ( f_x ) between ( x = 0.4 ) and ( x = 0.8 ) at ( y = 0.4 ):
[
f_{xx}(0.6, 0.4) \approx \frac{f_x(0.8, 0.4) – f_x(0.4, 0.4)}{0.8 – 0.4} = \frac{110 – 110}{0.4} = 0
]
Thus, ( f_{xx}(0.6, 0.4) \approx 0 ).
- Estimating ( f_{xy}(0.6, 0.4) ):
- The mixed partial derivative ( f_{xy} ) represents the rate of change of ( f_x ) with respect to ( y ). To estimate ( f_{xy} ) at ( (0.6, 0.4) ), consider the change in ( f_x ) between ( y = 0.2 ) and ( y = 0.6 ) at ( x = 0.6 ):
[
f_{xy}(0.6, 0.4) \approx \frac{f_x(0.6, 0.6) – f_x(0.6, 0.2)}{0.6 – 0.2} = \frac{110 – 110}{0.4} = 0
]