LOCAL LINEAR APPROXIMATION MATH 200 GOALS Be able to compute the - PowerPoint PPT Presentation

MATH 200 WEEK 6 - WEDNESDAY LOCAL LINEAR APPROXIMATION

MATH 200 GOALS ▸ Be able to compute the local linear approximation for a function of two or more variables at a given point. ▸ Be able to use a local linear approximation to estimate a given quantity.

MATH 200 FROM CALC 1 ▸ In Calc 1 we discussed the fact that differentiable functions are locally linear ▸ That is, near the point of tangency , a function is approximately equal to its tangent line ▸ Another way of saying all this is that if I keep zooming in on a differentiable function at a point, it will eventually look flat .

MATH 200 ▸ Say we want to approximate sin( π /15) ▸ We can find the tangent line to f(x) = sin(x) at x=0, ZOOM IN which is close to π /15 ▸ f’(x) = cos(x) ▸ f’(0) = 1 ▸ The tangent to f at x=0 is y=x (call this L(x) = x) ▸ sin( π /15) is approximately equal to L( π /15) = π /15

MATH 200 SUMMARY OF CALC 1 STUFF ▸ Local Linear Approximation for single variable functions says that a differentiable function can be approximated by its tangent line ▸ For a differentiable function f(x), the local linear approximation at x = x 0 is given by ▸ L(x) = f(x 0 ) + f’(x 0 )(x - x 0 ) ▸ Remember : Don’t think of this a formula to be memorized; this is just the tangent line to f at x 0 !

MATH 200 NEW STUFF ▸ If a single-variable, differentiable function can be approximated by its tangent line near the point of tangency, then a multi-variable function, f(x,y), can be approximated by its tangent plane near the point of tangency

MATH 200 REVISITING TANGENT PLANES FOR FUNCTIONS OF TWO VARIABLES ▸ Consider any function of two variables, f(x,y). ▸ To find the tangent plane at (x 0 ,y 0 ), we should treat the surface z = f(x,y) as a level surface of some function of three variables: ▸ z = f(x,y) can be written as 0 = f(x,y) - z ▸ F(x,y,z) = f(x,y) - z ▸ Notice that F x = f x , F y = f y , and F z = -1 − → ▸ So, ∇ F = ⟨ f x , f y , − 1 ⟩ ▸ And this will be the case for any function of two variables!

MATH 200 ▸ We can use this to write a general formula for the tangent plane to f(x,y) at (x 0 ,y 0 ): f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) − ( z − z 0 ) = 0 ▸ Solve for z: z = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + z 0 ▸ Since z 0 = f(x 0 ,y 0 ), z = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + f ( x 0 , y 0 ) THE ONLY THING WE’LL DO DIFFERENTLY NOW IS RENAME Z Z = L(X,Y) L ( x, y ) = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + f ( x 0 , y 0 )

MATH 200 DON’T MEMORIZE, UNDERSTAND ▸ Now, we have this formula for the local linear approximation of a function f(x,y) at (x 0 ,y 0 ): L ( x, y ) = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + f ( x 0 , y 0 ) ▸ But, it’s most important to remember that we approximate functions of two variables with tangent planes ▸ And we know that the normal vector for a tangent plane comes from the gradient ▸ You should be able to derive this formula if you forget it

MATH 200 EXAMPLE 1 ▸ Following our newfound ▸ Consider the function formula, we need f(0,0), f(x,y) = e y sin(x). Use local f x (0,0), and f y (0,0) linear approximation to approximate the value of ▸ f x (x,y) = e y cos(x); f x (0,0) = 1 f(0.1,0.1) ▸ f y (x,y) = e y sin(x); f y (0,0) = 0 ▸ We can evaluate the function f at (0,0) , ▸ f(0,0) = 0 which is close to (0.1,0.1), so we’ll pick that as the point of tangency .

MATH 200 ▸ Putting it all together… ▸ f x (x,y) = e y cos(x); f x (0,0) = 1 ▸ f y (x,y) = e y sin(x); f y (0,0) = 0 ▸ f(0,0) = 0 L ( x, y ) = f x ( x 0 , y 0 )( x − x 0 ) + f y ( x 0 , y 0 )( y − y 0 ) + f ( x 0 , y 0 ) L ( x, y ) = (1)( x − 0) + (0)( y − 0) + 0 L ( x, y ) = x SO BASICALLY 0.1 ▸ So… ▸ Wolfram vs. Us 0.1103329887302037 f (0 . 1 , 0 . 1) = L (0 . 1 , 0 . 1) e 0 . 1 sin(0 . 1) = 117193358278087139 e 0 . 1 sin(0 . 1) = 0 . 1 888318352848859486

So, the z-value at (0.1,0.1) for the surface, is really close to the z-value at (0.1,0.1) on the plane

MATH 200 EXAMPLE 2 ▸ Find the local linear approximation, L(x,y), for √ f ( x, y ) = ln( x 2 − y 2 ) at P (2 , 3) 2 x 3) = 4 √ f x ( x, y ) = x 2 − y 2 = ⇒ f x (2 , 1 = 4 √ − 2 y 3) = − 2 3 √ √ f y ( x, y ) = x 2 − y 2 = ⇒ f y (2 , = − 2 3 1 √ f (2 , 3) = ln(1) = 0 √ √ L ( x, y ) = 4( x − 2) − 2 3( y − 3) + 0 √ L ( x, y ) = 4 x − 2 3 y − 2

MATH 200