MATH 150 SUPPLEMENTAL NOTES 16

LINEARIZATION AND DIFFERENTIALS

LINEARIZATION

 Consider the function y = x 2 + 4. Find its tangent line at the point (1, 5).
 y' = 2x ® m = 2(1) = 2 y - 5 = 2(x - 1) ® y - 5 = 2x - 2 ® y = 2x + 3 Now, let us look at the graph of the function and its tangent line. Notice that as you get close to the point (1, 5), the function y = x 2 + 4 starts to look like y = 2x + 3. In fact, we can use the tangent line to approximate a value of a function at a point. y = x 2 + 4 y = 2x + 3
 DEFINITION: If f is differentiable at x = a, then the approximating function L (x) = f (a) + f ' (a)(x - a) is the linearization of f at a.
 EXAMPLE 1: Find the linearization of f (x) = x - 1 at x = 2.
 SOLUTION: First evaluate the function at x = 2, and then find the first derivative and evaluate it at x = 2. Now plug the information into the formula for the linearization.
 EXAMPLE 2: Find the linearization of f (x) = x 3 - 2x + 3 at x = 2.
 SOLUTION: First, evaluate the function at x = 2. After that, find f ' (x) and evaluate it at x = 2. Now plug the information into the formula for the linearization. L (x) = 7 + 10(x - 2) ® L (x) = 7 + 10x - 20 ® L (x) = 10x - 13
 EXAMPLE 3: Find a linearization of f (x) = 2x 2 + 4x - 3 that will approximate x 0 = -0.9 reasonably well.
 SOLUTION: I want to pick a value for a that is close to -0.9 and easy to work with, so I will pick a = -1.
 EXAMPLE 4: Find a linearization of f (x) = x /(x + 1) that will approximate x 0 = 1.3 reasonably well.
 SOLUTION: A value that is close to 1.3 is 1, so I will let a = 1.

DIFFERENTIALS

 DEFINITION: Let y = f (x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy = f ' (x) dx.
 EXAMPLE 5:
 SOLUTION: To find the differential, we will first find dy/dx and solve it for dy in terms of x and dx.
 EXAMPLE 6: Find dy for y = sec (x 2 - 1).
 SOLUTION: Why do we want to learn how to find a differential? We can use a differential to estimate the change a function makes as we move from x 0 to a point nearby. Sometimes this is used as an error measurement in approximation methods. The key here is that it is an estimated change. So not let us talk about the three types of change - Absolute change, Relative change, and Percentage change. I will state both the true and estimated formulas for each.
 TRUE ESTIMATED ABSOLUTE CHANGE D f = f (x 0 + dx) - f (x 0) df = f ' (x 0) dx RELATIVE CHANGE PRECENTAGE CHANGE
 We can use these three ways to describe the change in f. So let us do a couple of examples to illustrate these types of change.
 EXAMPLE 7: Write a differential formula that estimates the change in the surface area S = 6x 2 of a cube when the edge lengths change from x 0 to x 0 + dx.
 SOLUTION: First, find the differential dS. Now plug in x 0 to find the change in S when x moves from x 0 to x 0 + dx. dS = 12x 0 dx
 EXAMPLE 8: The radius of a circle is increased from 2.00 to 2.02 m. (a) Estimate the resulting change in area, and (b) express the estimate in (a) as a percentage of the circle's original area.
 SOLUTION: (a) The formula for the area of a circle is A = p r 2. The starting radius is 2.00 m and the change dr = 0.02 m. We want to find dA in this case. (b) To find the percentage change, I will divide dA by the area of the circle with radius of 2.00 m. Then I will multiply this result by 100.
 EXAMPLE 9: The edge of a cube is measured as 10 cm with an error of 1%. The cube's volume is to be calculated from this measurement. Estimate the percentage error in the volume calculation.
 SOLUTION: The formula for the volume of a cube is V = x 3. x = 10 cm and dx = 0.01.

As you work through the examples set forth in this set of supplemental notes, make sure that you understand the concept of linearization. We will revisit this topic in calculus II, so if you master it now, it will be easier later. Also, work through the differential examples. If you have any questions on any of these examples, please feel free to contact me.