draw a new location of the pole barn on the graph below. Note that the stock tank is located at the origin.
1 answer:
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Answer:
Step-by-step explanation:
First, let's change those variables to x and y, just for the sake of convenience. In order to find the inverse of a function algebraically, switch the x and y coordinates, then solve for the new y. Letting y = A(n) and x = n (we will switch them back when we're done):
y = 3x - 20. This is linear; a line with a slope of 3 and a y-intercept of -20. When we switch the x and the y, we get:
x = 3y - 20. Now we solve for the new y. Begin by adding 20 to both sides:
x + 20 = 3y. Now divide both sides by 3:
, or to write it in slope-intercept form, like the function you started with:
This is also a line, with a slope of 1/3 and a y-intercept of +20/3
Now, replacing:
That is how to write the inverse using function notation. The little -1 as an exponent tells us that this is the inverse of the function A(n).
Answer:
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Approximation at points (1.1,1.9) is 0.7
Step-by-step explanation:
Given:
Tangent plane to a surface z=5x+2y-10 as the function at point (1,2)
To find :
f(x,y) at (1,2)
partial derivatives of function w.r.t. (x and y) and value of that function at given points.
Solution:(refer the attachment also)
Now we know that
the equation of tangent plane at given points to the surface is given by,
f(x1,y1,z1) and z=f(x,y)
z-z1=Fx(x1,y1)*(x-x1)+Fy(x1,y1)*(y-y1)
here Fx(x1,y1) and Fy(x1,y1) are the partial derivatives of x and y.
now
taking partial derivative w.r.t. x we get
Fx(x1`,y1)=
=5.
Then w.r.t y we get
Fy(x1,y1)=
=2.
The values for Fx(1,2) and Fy(1,2) are 5 and 2 respectively.
Using the Linearization or linear approximation we get
L(x,y)=f(x1,y1)+Fx(x,y)*(x-x1)+Fy(x,y)(y-y1)
=-1+5(x-1)+2(y-2)
=5x+2y-10
Approximation at F(1.1,1.9)
=5(1.1)+2(1.9)-10
=5.5+3.8-10
=0.7
Approximation at points (1.1,1.9) is 0.7
