3.7 applications of tangent lines

Applications of Tangent Lines


In this section we look at two applications of the

tangent lines.



tangent lines.

Differentials and Linear Approximation



tangent lines.


Let f(x) = x2 be as shown. y

Its derivative f '(x) = = 2x. dydx

y = x2

x



tangent lines.



Its derivative f '(x) = = 2x.

The slope at the point (3,9)

is f '(3) = 6.

dydx

(3,9)

y = x2

x



tangent lines.





is f '(3) = 6.

dydx

(3,9)

y = x2y = 6x – 9

Hence the tangent line at (3, 9) is

y = 6(x – 3) + 9 or y = 6x – 9.x



tangent lines.





is f '(3) = 6.

dydx

(3,9)

y = x2y = 6x – 9

Hence the tangent line at (3, 9) is

y = 6(x – 3) + 9 or y = 6x – 9.

Note that in the notation we write

that

dxdydx

x = 3

= 6dydx

x = a

= f '(a)

dy

where in general

x

In general the tangent line

at x = a, i.e. at (a, f(a)), is

with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)




with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)


point x = a, say x = b, the

function–values f(b) and T(b) are

near each other,

For any x value that is near the

(a,f(a))

T(x) = y

(b,f(b))

(b,T(b))

y = f(x)



with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)




near each other,


(a,f(a))

T(x) = y

(b,f(b))

(b,T(b))

y = f(x)f(b)



with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)




near each other,


(a,f(a))

T(x) = y

(b,f(b))

(b,T(b))

y = f(x)T(b) f(b)



with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)




near each other, hence we may

use T(b) as an estimation for f(b),


(a,f(a))

T(x) = y

(b,f(b))

(b,T(b))

y = f(x)T(b) f(b)



with f'(a) = dydx

x = a

T(x) = f'(a)(x – a) + f(a)

(a,f(a))

y = f(x)

x

tangent line

T(x) = f'(a)(x – a) + f(a)




near each other, hence we may

use T(b) as an estimation for f(b),

since T(x) is a linear function

and most likely it’s easier to

calculate T(b) than f(b).


(a,f(a))

T(x) = y

(b,f(b))

(b,T(b))

y = f(x)T(b) f(b)


There are two ways to find T(b).



1. Find T(x) and evaluate T(b) directly.



1. Find T(x) and evaluate T(b) directly. For example,

the tangent line of y = sin(x) at x = 0 is T(x) = x,





hence for small x’s that are near 0, sin(x) ≈ x.

(Remember that x must be in radian measurements).







(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a x=b

2. Find ΔT first and ΔT + f(a) = T(b).







(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

ΔT

x=b








which is identified with the dy in dy/dx.

(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

ΔT= dy

x=b


In practice, ΔT is denoted as dy, the y–differential,








And we set the x–differential dx = Δx,

(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b











and as before Δy = f(b) – f(a). (a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b



Δy









and as before Δy = f(b) – f(a).

These measurements are

shown here. It’s important to

(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy



“see” them because the

geometry is linked to the algebra.


(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy

The derivative notation is inherited from the

notation of slopes as ratios.

dydx

ΔyΔx


Δx0= L (= lim ),

(a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy

dydx

x = a

= f '(a) ΔyΔx


notation of slopes as ratios. Suppose that

dydx

ΔyΔx


Δx0= L (= lim ),

ΔyΔx

then for Δx close to 0, (a,f(a))

(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy

dydx

x = a

= f '(a) ΔyΔx

≈ L = slope at (a, f(a))



dydx

ΔyΔx


Δx0= L (= lim ),

ΔyΔx


(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy

Δy ≈ L(Δx)

dydx

x = a

= f '(a) ΔyΔx

≈ L = slope at (a, f(a)) so



dydx

ΔyΔx


Δx0= L (= lim ),

ΔyΔx


(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy

Δy ≈ L(Δx) = L dx = dy

dydx

x = a

= f '(a) ΔyΔx




dydx

ΔyΔx


Δx0= L (= lim ),

ΔyΔx


(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy


dydx

x = a

= f '(a) ΔyΔx

The accuracy of this approximation

may be formulated mathematically.




dydx

ΔyΔx



Applications of Tangent Linesdydx

ΔyΔx

Δx0= L (= lim ),

ΔyΔx



(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy


dydx

x = a

= f '(a) ΔyΔx

So what have we done?

We have clarified the meaning of dy and dx, and

made it legal to treat dy/dx as a fraction both

algebraically and geometrically.



Applications of Tangent Linesdydx

ΔyΔx

Δx0= L (= lim ),

ΔyΔx



(x. T(x))

(b,f(b))

(b,T(b))

y = f(x)

x=a

Δx=dx

ΔT= dy

x=b

Δy


dydx

x = a

= f '(a) ΔyΔx

So what have we done?

We have clarified the meaning of dy and dx, and

made it legal to treat dy/dx as a fraction both

algebraically and geometrically. To summarize, given dydx

= f '(x), then Δy ≈ dy = f '(x)dx.y = f(x) and


Example A. Let y = f(x) = √x.

a. Find the general formula of dy in terms of dx.



a. Find the general formula of dy in terms of dx.dydx = f '(x) = ½ x–½



a. Find the general formula of dy in terms of dx.dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx




b. Find the specific dy in terms of dx when x = 4.

dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx





dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx

When x = 4 we get that dy = ¼ dx.





dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx

When x = 4 we get that dy = ¼ dx. (This says that for

small changes in x, the change in the output y is

approximate ¼ of the given change in x, at x = 4.)





c. Given that Δx = dx = 0.01, find dy at x = 4.

Use the result to approximate √4.01.

dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx










dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx




Given that Δx = dx = 0.01 and that dy = ¼ dx at x = 4,







dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx





we have dy = ¼ (0.01) = 0.0025 ≈ Δy.







dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx





we have dy = ¼ (0.01) = 0.0025 ≈ Δy.

Hence f(4 + 0.01) =√4.01 ≈ √4 + dy







dydx = f '(x) = ½ x–½ or dy = ½ x–½ dx





we have dy = ¼ (0.01) = 0.0025 ≈ Δy.

Hence f(4 + 0.01) =√4.01 ≈ √4 + dy = 4.0025.

(The calculator answer is 002498439..).

Your turn. Do the same at x = 9, and use the result to

approximate √8.995. What is the dx?





Hence we have the terms the “differentials” or

the “small differences” as opposed to the “differences”






and the symbols are dx and Δx respectively.







The differentials are used extensively in numerical

problems and the following is one example.









Newton’s Method for Approximating Roots










The Newton’s Method of approximating the location

of a root depends on the geometry of the tangent

line and the x–axis.










The Newton’s Method of approximating the location

of a root depends on the geometry of the tangent

line and the x–axis. If the geometry is right,

the successive x–intercepts of tangent–lines

approach a root not unlike a ball falls into a crevice.

We demonstrate this below.


Suppose that we know there is a root in the shaded

region and we wish to calculate its location, and let’s

say it’s at x = r.

y = f(x)

x = r




say it’s at x = r. We select a starting point x = x1 as

shown.

y = f(x)

x = r

x1





shown. Draw the tangent line at x1.

(x1,f(x1))

y = f(x)

x = r

x1





shown. Draw the tangent line at x1. Let x2 be the

x–intercept of this tangent.

(x1,f(x1))

y = f(x)

x = r

x1x2






x–intercept of this tangent. Then we draw the tangent

line at x2, and let x3 be the x–intercept of the tangent

line at x2.

(x1,f(x1))

y = f(x)

x = r

x1x3 x2

(x2,f(x2))






x–intercept of this tangent. Then we draw the tangent

line at x2, and let x3 be the x–intercept of the tangent

line at x2. Continuing in this manner, the sequence of

x’s is funneled toward the root x = r.

(x1,f(x1))

y = f(x)

x = r

x1x3 x2

(x3,f(x3))

(x2,f(x2))


The successive intercepts may be calculated easily.

We give the formula here:

xn+1 = xn – f(xn)f '(xn)


The successive intercepts may be calculated easily.

We give the formula here:

Example B.

Let y = f(x) = x3 – 3x2 – 5, its

graph is shown here. Assume

that we know that it has a root

between 2 < x < 5.

Starting with x1 = 4,

approximate this roots by the

Newton’s method to x3.

Compare this with a calculator

answer.

xn+1 = xn – f(xn)f '(xn)

2 4

y

x

f(x) = x3 – 3x2 – 5


We have that f '(x) = 3x2 – 6x = 3x(x – 2)


x1=42

x

f(x) = x3 – 3x2 – 5

The geometry of the Newton’s Method

for example B.

Back to math–265 pg

http://www.lahc.edu/math/cal2/math_265.html


x2 = x1 – f(x1)f '(x1)



x1=42

x

f(x) = x3 – 3x2 – 5


for example B.


x2



x2 = x1 – f(x1)f '(x1)



= 4 – f(4)f '(4)

x1=42

x

f(x) = x3 – 3x2 – 5


for example B.


x2



x2 = x1 – f(x1)f '(x1)



= 4 – f(4)f '(4)

= 85/24

x1=42

x

f(x) = x3 – 3x2 – 5

≈ 3.542


for example B.


x2



x2 = x1 – f(x1)f '(x1)



= 4 – f(4)f '(4)

= 85/24

x3 = 85/24 –f(85/24)f '(85/24)

x1=42

x

f(x) = x3 – 3x2 – 5

≈ 3.542

x3 x2


for example B.




x2 = x1 – f(x1)f '(x1)



= 4 – f(4)f '(4)

= 85/24

x3 = 85/24 –f(85/24)f '(85/24)

≈ 3.432x1=42

x

f(x) = x3 – 3x2 – 5

≈ 3.542

x3 x2


for example B.




x2 = x1 – f(x1)f '(x1)



= 4 – f(4)f '(4)

= 85/24

x3 = 85/24 –f(85/24)f '(85/24)

≈ 3.432x1=42

x

f(x) = x3 – 3x2 – 5

≈ 3.542

x3 x2The software answer is

≈ 3.425988..


for example B.



3.7 applications of tangent lines

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