1. Linear Approximation and Differentials
Lesson 3.9

2. Tangent Line Approximation
Consider a tangent to a function at a point where x = a
Close to the point, the tangent line is an approximation for f(x)
y=f(x)
The equation of the tangent line: y = f(a) + f ‘(a)(x – a)
f(a) • a

3. Tangent Line Approximation
We claim that
This is called linearization of the function at x=a.
Recall that when we zoom in on an interval of a differentiable function far enough, it looks like a line.

4. New Look at dy = rise of tangent relative to x = dx• • • x
x + x
x = dx
dy = rise of tangent relative to x = dx
y = change in y that occurs relative to x = dx

5. New Look at We know that Recall that dy/dx is NOT a quotient
then
Recall that dy/dx is NOT a quotient
it is the notation for the derivative
However … sometimes it is useful to use dy and dx as actual quantities

6. The Differential of y Consider Then we can say
this is called the differential of y
the notation is d(f(x)) = f ’(x) * dx
it is an approximation of the actual change of y for a small change of x

7. Animated Graphical View
Note how the "del y" and the dy in the figure get closer and closer

8. Differentials Example: Find the differential of the function:
Remember a differential is dy = f ‘(x)· dx (assuming y=f(x))

9. Try It Out Note the examples and rules for differentials on page 238.
Find the differential of: 1) y = 3 – 5x2 2) f(x) = xe-2x

10. Differentials Example Finding the Differential dy
Find the differential dy and evaluate dy for the given value of x and dx.

11. Differential Estimate of Change (needed for #34 and #37 on the homework)
Three types of changes that can be found Absolute, Relative and Percent Change
Absolute
True
Estimated
Relative and Percent Change
True Estimate
Relative
Percent % %

12. <- Approximate error in volume
Differentials
Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume.
To solve this problem, we will use the equation:
∆V ≈ dV = V ‘(r)· dr
<- Approximate error in volume

13. Differential
Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume.
The estimate of relative error is given by:
<- Approximate relative error in volume
The estimate percentage of
relative error is given by:
<- Approximate relative error percentage in volume

14. Differentials for Approximations
Consider
Use
Then with x = 25, dx = .3 obtain approximation

15. Propagated Error Consider a rectangular box with a square base
Height is 2 times length of sides of base
Given that x = 3.5
You are able to measure with 3% accuracy
What is the error propagated for the volume? x 2x x

16. Propagated Error We know that
Then dy = 6x2 dx = 6 * 3.52 * = This is the approximate propagated error for the volume

17. Propagated Error The propagated error is the dy The relative error is
sometimes called the df
The relative error is
The percentage of error
relative error * 100% (in this case, it would be 9%)