Calc1131W5T3 - Linear approximation
Friday, 30 March 2018
12:28 AM
Created with Microsoft OneNote 2016.
![Question One
Untitled picture.png If you zoom in on the graph of a differentiable function f@) at the point [a, f(a)], then eventually it looks like a straight line. For example, if you look really closely at the function
f@) = near the point [16, 4] then it looks like
4.03
4.02
4.01
4
3.99
3.98
3.97
15.7
The tangent to f@) at [a, f(a)] is the line
15.9
16
16.1
16.2
163
y — f(a) = f' — a)
Since
the tangent to f(x)
at [16, 4] is given by the equation
Untitled picture.png Function values such as 15.8 are difficult to calculate exactly. But we can use the tangent to estimate the function for values near 16
(15.8) = 3.975000000..
This estimate is pretty close to
15.8 = 3.974921383 • • •
However, using the tangent at 16 is not accurate for x values far away from 16. For example,
. but
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](Calc1131W5T3%20-%20Linear%20approximation_files/image001.png)
![Untitled picture.png Some values of sin(x) are familiar and easy to evaulate; such as
5
But what about sin + 0.07 ? Exact computation is problematic; but the tangent at x
of sin@) in red, and the tangent at
in dotted blue line.
0.4
0.2
gives a close estimate to the function when x is near
The picture below shows the graph
0.2
(x) when
• Evaluate the derivative of sin
• Hence the equation of the tangent to sin(x) at
• So we may approximate sin
7T + 0.07 by evaluating I(x) at
0.4
* (x-pii6) + 1/2
+ 0.07
Untitled picture.png This gives
which is a pretty good approximation since:
Note: the Maple syntax for v,'7r is sqrt (Pi)
+ 0.07
+ 0.07
.5606217783 • • •
.5593477829 • • •
Question Three
Untitled picture.png Machine generated alternative text:
But we can find a better approximatiom Since
f' (256)
the tangent line to y = f(x) at the point [256, 4] is
This can be expressed as
So a better approximation is (to 4 decimal places)
and the actual value is:
1
Note: the Maple syntax for or x 4
Suppose that f@) = We can roughly expect that
f(256.3) f(256) ¯
(1/4)
y -4 = — 256)
4
256.3 = 4.001171360 • •
Question Four
Untitled picture.png Machine generated alternative text:
Consider the function y
The derivative is
ln@) _ Use the a-slider to see the tangent to y = f@) at x
e = 47
y — In(4.7) =
4.7
A = [4.7, In(4.7)]
The tangent line to at [1, O] is
The tangent line to at [2, In(2)] is
The tangent line to at [3, In(3)] is
Y = C3(x) —
x,'3-1+Ir13
Using the slider, or numerical calculation, we deduce that the best approximation to ln@) at x = 1.1 is given by
@ 11 (1.1)
0 12(1.1)
0 13(1.1)
Note: the Maple syntax for In(4) is In (4)
Find the derivative of the function at a point (we want to use this point as our reference)
Use the point-gradient form of a line to find the equation of the tangent.
Rearrange to give 𝑦, which is also equal to 𝑙𝑥
](Calc1131W5T3%20-%20Linear%20approximation_files/image003.png)
![Untitled picture.png Machine generated alternative text:
Consider the function y
The derivative is
ln@) _ Use the a-slider to see the tangent to y = f@) at x
e = 47
y — In(4.7) =
4.7
A = [4.7, In(4.7)]
The tangent line to at [1, O] is
The tangent line to at [2, In(2)] is
The tangent line to at [3, In(3)] is
Y = C3(x) —
x,'3-1+Ir13
Using the slider, or numerical calculation, we deduce that the best approximation to ln@) at x = 1.1 is given by
@ 11 (1.1)
0 12(1.1)
0 13(1.1)
Note: the Maple syntax for In(4) is In (4)
Question Five
Untitled picture.png Machine generated alternative text:
Tangent line approximations are also valid for more general curves. Consider for example the cubic curve given by the relation
-2
2
3
4
This relation can not be expressed so easily as a function y
dy
To find — we use implicit differentiation: this gives
where A
O Thus we can find that
dy
-2
f(x), and so finding the y-value for a given am-value generally requires approximation of some kind.
dy
dc
dy
dc
At the point [2, —1], the derivative has the value
dc
So we can approximate f(2.002) to 4 decimal places with
-5/12
O , so that the equation of the tangent line at [2, —1] is
-5X,'12 - 2/12
f(2.002) 1(2.002) —](Calc1131W5T3%20-%20Linear%20approximation_files/image004.png)
![Untitled picture.png Machine generated alternative text:
Tangent line approximations are also valid for more general curves. Consider for example the cubic curve given by the relation
-2
2
3
4
This relation can not be expressed so easily as a function y
dy
To find — we use implicit differentiation: this gives
where A
O Thus we can find that
dy
-2
f(x), and so finding the y-value for a given am-value generally requires approximation of some kind.
dy
dc
dy
dc
At the point [2, —1], the derivative has the value
dc
So we can approximate f(2.002) to 4 decimal places with
-5/12
O , so that the equation of the tangent line at [2, —1] is
-5X,'12 - 2/12
f(2.002) 1(2.002) —](Calc1131W5T3%20-%20Linear%20approximation_files/image005.png)