Linear Approximations  Differentials  Links to Other Explanations of Differentials  Check Concepts
Review: Linear Approximations were first experienced in Lesson 2.9. It would be healthy to go back and briefly review our first contact with this topic.
In the two graphs above, we are reminded of the principle that
a tangent line to a curve at a certain point can be a good approximation of
the value of a function if we are "close by" the point we are interested
in. In the above graphs, we see that near the point (8,2) (the point of tangency)
the green tangent is extremely close to
the red original curve. We also notice that
the closer we get to the point of tangency, the more accurate our linear approximation
is.
If you have the Journey Through Calculus CD, load and run MResources/Module 3/Linear Approximations/Start of Linear Approximations. 
Example problem: Find a linear approximation or linearization for . Use this approximation to estimate .
We first calculate the derivative of the function. This will allow us to know the general formula for the slope of the tangent line at any point on the curve.  
Next, we need to find the slope of the tangent line at x = 5.  
Now we can begin to formulate the equation of the tangent line, because we know its specific slope.  
We now subsitute in the coordinates of the given point or (5,1.71) and solve for b.  
We now have our linear approximation.  
Now we can find our estimate for the cube root of 5.03 
If we take the two derivative notations that we have been using and set them equal, we have the equation: . If we then multiply both sides of this equation by we get: . This equation shows that we can calculate dy as a dependent variable, based on the inputs of dependent variables x and dx.This means that we can calculate an estimated error (dy) in linear approximation using such a formula. An example follows.
Sample Problem: The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere? 
If the radius of the sphere is r then its
volume is .
If the error in the measured value of r is denoted by then
the corresponding error in the calculated value of V is ,
which can be approximated by the differential . 
When r = 21 and dr = 0.05, this becomes: 
Therefore, the maximum error in the calculated volume is about 277 cubic centimeters. 
Links to Other Explanations
of Differentials:
Differentials
University of Kentucky Tutorial on Differentials
If you have the Journey Through Calculus CD, load and run MResources/Module 3/Linear Approximation/Start of Linear Approximation. 
