2.13 Linearization and Differentials
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1 Linearization and Differentials Section Notes Page Sometimes we can approimate more complicated functions with simpler ones These would give us enough accuracy for certain situations and are easier to work with The approimating functions in this section are called linearizations, and they are based on tangent lines It is possible to have approimating functions that are polynomials, but we will not discuss this here Given the following figure, we want to come up with and equation for our tangent line As we zoom in, the closer the tangent line will lay on top of the original curve allowing us to find the eact slope at the value of = a In order to find the equation for the tangent line we will start with the point-slope formula y y m( ) Here the m (slope) is f (a) Also a and y f ( a) We will substitute both of these into the point-slope formula to get y f ( a) f ( a) We will now solve for y y f ( a) f ( a) The y is epressed as L() f ( a) f ( a) This is the linearization of f at a If f is differentiable at = a, then the approimating function is f ( a) f ( a) is the linearization of f at a The approimation f ( ) L( ) of f by L is the standard linearization of f at a The point = a is the center of the linearization EXAMPLE Find the linearization L() of f ( ) 6 at = So we will be using f ( a) f ( a) So we need to find the individual components of this formula and substitute it all in Let s first find f(a) f () ( ) Then we need to find f (a) In order to do this we will first we will find the derivative f ( ) 6 () This can be simplified to f ( ) Now we will find f (a) f ( ) Now we will substitute everything 6 ( ) 6 into the linearization formula f ( a) f ( a) ( ( )) ( ) 9 6 Simplifying gives us which is our final answer
2 EXAMPLE Find the linearization L() of f ( ) sin at 6 Section Notes Page So we will be using f ( a) f ( a) So we need to find the individual components of this formula and substitute it all in Let s first find f(a) f sin Then we need to find f (a) In order to do 6 6 this we will first we will find the derivative f ( ) cos Now we will find f (a) f cos 6 6 Now we will substitute everything into the linearization formula f ( a) f ( a) 6 6 EXAMPLE Given 9, find a linearization that will replace f ( ) 4 by centering your linearization not on, but at a nearby integer = a This will make the derivative and function easy to evaluate We want to find an integer that is closest to, so in this case we will let a We will now use f ( a) f ( a) So we need to find the individual components of this formula and substitute it all in Let s first find f(a) f ( ) ( ) Then we need to find f (a) In order to do this we will first we will find the derivative f ( ) 6 Now we will find f (a) f 6( ) 4 Now we will substitute everything into the linearization formula f ( a) f ( a) 4 ( ) 4( ) EXAMPLE Given, find a linearization that will replace f ( ) by centering your linearization not on, but at a nearby integer = a This will make the derivative and function easy to evaluate We want to find an integer that is closest to, so in this case we will let a We will now use f ( a) f ( a) So we need to find the individual components of this formula and substitute it all
3 Section Notes Page () in Let s first find f(a) f ( ) 4 Then we need to find f (a) In order to do this we will first we will ( )() () find the derivative This will involve the quotient rule f ( ) ( ) ( ) ( ) Now we will find f (a) f Now we will substitute everything into the ( ) linearization formula f ( a) f ( a) 4 ) EXAMPLE Given, find a linearization that will replace f ( ) tan by centering your linearization not on, but at a nearby integer = a This will make the derivative and function easy to evaluate We want to find an integer that is closest to, so in this case we will let a We will now use f ( a) f ( a) So we need to find the individual components of this formula and substitute it all in Let s first find f(a) f tan Then we need to find f (a) In order to do this we will first we will find the derivative f ( ) Now we will find f (a) f Now we will substitute everything into the linearization formula f ( a) f ( a) This is our final answer Differentials Back in the section regarding implicit differentiation, we were solving for We treated this as one variable However in this section we will consider this made up of two different variables, and Let y f () be a differentiable function The differential is an independent variable The differential is f ( ) But what does this mean in terms of something physical? The below diagram illustrates this geometrically Let = a and set The corresponding change in y f () is y f ( a ) f ( a) The corresponding change in the tangent line L is L L( a ) L( a) f ( a) f ( a)[( a ) a] f ( a) f ( a) Therefore the change in the linearization of f is precisely the value of the differential when = a and So the represents the amount the tangent line rises or falls when changes by an amount
4 EXAMPLE Find given y 4 6 Section Notes Page 4 First we will find the derivative We are using this notation since we are working with differentials So, 4 6 This simplifies to 4 The and can be thought of as two separate variables, so multiply both sides by in order to isolate the You will get the final answer, which is 4 EXAMPLE Find given y y 8 This one involves implicit differentiation The first term y involves the product rule y y () 8 Now simplify y y Get all terms with on one side of the equation y y Factor out the y y Now solve for y y Now multiply both sides by y y It is now solved for, so this is our answer EXAMPLE Estimate the volume of material in a cone-shaped shell with a height of in, radius 9 in, and shell thickness in First we start with the formula for a cone V r h Then we want to take the derivative with respect to r dv rh Notice that since we are taking the derivative with respect to r, the h is treated like a constant dr Now we will multiply both sides by dr dv rhdr Now we plug in our given information We are given that h = in, r = 9 in, and dr = in dv (9)() () dv in
5 Section Notes Page EXAMPLE The radius of a sphere is measured as 6 cm with an error of % The volume of the sphere is to be calculated from this measurement Estimate the percentage error in the volume calculation To estimate the percentage error in the volume calculation, we will divide the estimated change in volume due to error by the actual volume of the sphere with the given radius First let s find the change in volume due to error This is the same as dv First we start with the formula for a sphere 4 V r Net we will take the derivative of both sides of our volume formula with respect to r dv 4 dv r This simplifies to 4r If we multiply both sides by dr we get dv 4r dr We need dr dr to know which information to plug into this To find dr, we will multiply the measurement of the radius by the error, which in this case is % So dr = 6() = 8 Now we can plug in the values to find dv dv 4 (6) (8) 9 cm 4 Net, let s find the actual volume of the sphere with the given radius V (6) 88 cm 9 So to find the percentage error in the volume calculation, divide 9 9% 88
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Sets 1.2 Find the set of all x R satisfying > > Test x 0, 5 > - > 0 [x- 3 (x -2)] > 0 - (x 1) (x 2) (x 3) > 0 At x0: y - (-1)(-2)(-3) 6 > 0 x < 1 At x5: y - (4)(3)(2) -24 < 0 2 < x < 3 Hence, {x R: x
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