Find the linear approximation of the function z=ln(x-3y) at the point (7, 2). Use the linear approximation to approximate f (6.9, 2.06). Compare the approximation to the actual value at that point. What is the relative error of the approximation there? (To find the relative error a decimal calculate the difference between the actual value and the approximation and then divide by the actual value. This can be converted into a percent.) Repeat this for f (6.8, 2.07).
What do you observe?

Question

Find the linear approximation of the function z=ln(x-3y) at the point (7, 2). Use the linear approximation to approximate f (6.9, 2.06). Compare the approximation to the actual value at that point. What is the relative error of the approximation there? (To find the relative error a decimal calculate the difference between the actual value and the approximation and then divide by the actual value. This can be converted into a percent.) Repeat this for f (6.8, 2.07).
What do you observe?

myininaya · Answer

$$f(x,y) \approx f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \text{ <--- linear approximation }$$

myininaya · Answer

Do you know how to use that formula?

anonymous · Answer

no

myininaya · Answer

Do you know how to take the partial of f with respect to x?

anonymous · Answer

no

myininaya · Answer

This is an example of how to take partials
Ok what if I give you $$f(x,y)=x^2y$$
$$f_x(x,y)=2xy$$
$$f_y(x,y)=x^2$$
So in f_x you see I treated everything like a constant except x in f(x,y) 
f_y , I treated everything like a constant except for y in f(x,y)

myininaya · Answer

You try taking the partial of our f in your problem with respect to x
Remember you will have to use chain rule

myininaya · Answer

Can I see what you did?

myininaya · Answer

You should know how to take derivative of this:
$$\ln(x-3)$$
so how do you find derivative of this?

myininaya · Answer

I'm asking for you to evaluate: $$\frac{d}{dx}(\ln(x-3))$$

anonymous · Answer

1/x-3

myininaya · Answer

ok so if you have $$f(x,y)=\ln(x-3y) => f_x(x,y)=\frac{1}{x-3y}$$

myininaya · Answer

You treat y like it was a constant
the derivative of inside is 1 when y is a constant right?

myininaya · Answer

Now you try f_y

anonymous · Answer

-3/x-3y

myininaya · Answer

that's right

myininaya · Answer

So we have $$f(x,y) \approx f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) $$
So (a,b)=(7,2)
wherever you see a put 7 and wherever you see b put 2

myininaya · Answer

What is f(7,2)
what is f_x(7,2)
what is f_y(7,2)

anonymous · Answer

What is f(7,2)=0
what is f_x(7,2)=1
what is f_y(7,2)-=3

myininaya · Answer

great and the last one I think you mean =-3

anonymous · Answer

yes

myininaya · Answer

$$f(x,y)\approx 0+1(x-7)-3(y-2)$$
and i replaced the a with 7 and the b with 2
now find f(6.9,2.06) using our approximation for f(x,y)
just replace x with 6.9 and y with 2.06 and simplify

anonymous · Answer

=x-3y-1
=(6.9)-3(2.06)-1
=-0.28

myininaya · Answer

ok yeah the approximation i got was also -.28

myininaya · Answer

let me know if you need anymore help with the rest of your problem