**Linear Approximations and Differentials**
**Linear Approximation** (or **Tangent Line Approximation**): To estimate the value of a function (which might be difficult or impossible to evaluate without electronic aids) near a certain known point, *x* =*a*, on its graph by substituting its tangent line (because the values of a linear function is very easy to compute) in its place. The price of such a simplification is that the estimate is usually good only for a small interval around the given point, that is, when *x* is close to *a*.
Hence, given a function *f* , and suppose (*x*_{0}, *y*_{0}) = (*a*, *f* (*a*)) is a known point on the graph of *f* . Then the line tangent to the graph *y* = *f* (*x*) at *a* has slope *m* = *f* ′(*a*). Therefore, the point-slope form of the equation of tangent line is
*y* − *y*_{0} = *m*(*x* − *x*_{0}) *y* − *f* (*a*) = *f* ′(*a*) (*x* − *a*)
Simplified to the slope-intercept form:
*y* = *f* (*a*) + *f* ′(*a*) (*x* − *a*)
That is, when *x* is near *a*_{ }, *f* (*x*) ≈ *f* (*a*) + *f* ′(*a*) (*x* − *a*).
The expression
*L*(*x*) = *f* (*a*) + *f* ′(*a*) (*x* − *a*)
is called the __linearization__ of *f* at *a*.
*Ex*. Use linear approximation to estimate the value of sin(1).
First we need a known value of sin(*x*) near *x* = 1. The best point to use is at the reference angle π/3 ≈ 1.047, since it is the angle closest to *x* = 1. Hence, we will use the point (*x*_{0}, *y*_{0}) = (π/3, /2).
Let *f* (*x*) = sin *x*, then *f* ′(*x*) = cos *x*.
The line tangent to *y* = sin *x* at *x* = π/3, therefore, has slope *m* = *f* ′(π/3) = cos(π/3) = 1/2, and passes through the point (π/3, /2). Its equation is, therefore,
Simplifying,
Hence, the linearization of *y* = sin *x* at *x* = π/3 is
At *x* = 1,
The actual value of sin(1) is 0.84147…, so the estimate is accurate within 0.001.
**Differential**: To estimate the change in *y*-value of a function given the change in *x*-value. Here, the change in *x*, ∆*x* = *dx*, is considered to be the independent variable. The differential *dy* is the dependent variable and a function of *dx*. The relation between *dx* and *dy* is given by
Therefore,
*dy* = *f* ′(*x*) *dx*
Here, the value of *x* is fixed at some number *a*, so it is actually a* *constant. The independent variable is *dx*, and *dy* is the dependent variable.
*Comment*: The differential is just another way of looking at the linear approximation. When using the differential, we estimate the *change* in the *y*-value, as the *x*-value deviates from a fixed number *a*, rather than the new *y*-value itself.
*Notation*: ∆*x* = *dx* is the change in *x*; ∆*y* = *f* (*a* + ∆*x*) − *f* (*a*) is the change in *y* (the __actual__ change of function’s values from *x* = *a* to *x* = *a* + ∆*x*); and the differential, *dy* = *f* ′(*x*) *dx*, is the __estimated__ change of function’s values from *x* = *a* to *x* = *a* + ∆*x*, according to the linear approximation of *f* at *a*. In other words, *dy* = *L*(*a* + ∆*x*) − *f* (*a*), where *L*(*x*) is the linearization of *f* at *a*.
To see the relationship between differential and linear approximation, recall that the linearization of *f* at *a* is *L*(*x*) = *f* (*a*) + *f* ′(*a*) (*x* − *a*). When *x* is moved from *a* to *a* + ∆*x*, the expression becomes
*L*(*a* + ∆*x*) = *f* (*a*) + *f* ′(*a*) ((*a* + ∆*x*) − *a*) = *f* (*a*) + *f* ′(*a*) ∆*x*
Since ∆*x* = *dx*, *L*(*a* + ∆*x*) = *f* (*a*) + *f* ′(*a*) *dx* and
*dy* = *L*(*a* + ∆*x*) − *f* (*a*) = (*f* (*a*) + *f* ′(*a*) *dx*) − *f* (*a*) = *f* ′(*a*) *dx*.
Therefore, for any fixed value *x*, and a variable value *dx*
*dy* = *f* ′(*x*) *dx*.
*Comment*: Since linear approximations work best for values close to the fixed point *a*, so do differential become better estimates if the deviation *dx* becomes smaller (that is, if *x* is close to *a*).
*Ex*. Compute ∆*y* and *dy* for *y* = for *x* = 1 and
(i.) ∆*x* = 1.
(ii.) ∆*x* = 4.
At the fixed point *x* = *a* = 1, *f* ′(1) = 1/2.
(i.) ∆*y* = *f* (*a* + ∆*x*) − *f* (*a*) = *f* (2) − *f* (1) = − 1 = 0.414213…
*dy* = *f* ′(1) *dx* = *f* ′(1) ∆*x* = (0.5)(1) = 0.5
Therefore, the linear approximation with base point *a* = 1 would give an estimate of *f* (1) + *dy* = 1.5 for the value of , a reasonable estimate compares to the actual value of 1.414213…
(ii.) ∆*y* = *f* (*a* + ∆*x*) − *f* (*a*) = *f* (5) − *f* (1) = − 1 = 1.236067…
*dy* = *f* ′(1) *dx* = *f* ′(1) ∆*x* = (0.5)(4) = 2
Therefore, the linear approximation with base point *a* = 1 would give an estimate of *f* (1) + *dy* = 3 for the value of (compares to the actual value 2.236067…), which is not too good an approximation. The *dx* in this case is 4, which is too large a deviation from the base point, *x* = 1, to give an accurate result. |