# How do you find the linear approximation L to f at the designated point P. compare the error in approximating f by L at the specified point Q with the distance between P and Q given #f(x,y) = 1/sqrt(x^2+y^2)#, P(4,3) and Q(3.92, 3.01)?

The tangent plane

where

The normal vector to

Giving

At

so the tangent plane reads

Given a point

and a point

Their distance is

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To find the linear approximation ( L ) to ( f ) at the designated point ( P ), first, compute the partial derivatives ( f_x ) and ( f_y ) of ( f ) at ( P ). Then, use the formula:

[ L(x,y) = f(P) + f_x(P)(x-x_P) + f_y(P)(y-y_P) ]

where ( (x_P, y_P) ) are the coordinates of point ( P ).

The error in approximating ( f ) by ( L ) at the specified point ( Q ) can be approximated by the absolute difference between ( f(Q) ) and ( L(Q) ).

To compare the error in approximating ( f ) by ( L ) at ( Q ) with the distance between ( P ) and ( Q ), compute the distance between ( P ) and ( Q ) using the distance formula:

[ \text{Distance} = \sqrt{(x_Q-x_P)^2 + (y_Q-y_P)^2} ]

Then, compare this distance with the error obtained above.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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