Applications on Derivatives

7/29/2019 Applications on Derivatives

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Mr . Sher if Yehia Al Maraghy https://twitter.com/Mr_Sheri f_yehia

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Calculus Chapter (3) Appl ications on Deri vatives-29-

first: Geometric applications second : Applications on related time rate

1 2 1 2

1 2 1

2

1 m Tan If is the ve direction of x Axis is given .

2 If two lines are parallel L // L Then m m

-13 If two lines are perpendicular L L then m " Normal slopes"

m

4 If the line is parallel to x axis to

y axis Then m 0

15 If the line is parallel to y axis to x axis then m

0

6 If the line intersects x - axis means y 0

7 If the line intersects y - axis means x 0

Applications on derivatives

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We know that The equation of the straight line : y m x c

WhereCoefficient of x

m The Slope =Coefficient of y

and C : the yintercept are given

And if a Slope(m) and a point 1 1x ,y are given , Then the equation of the straight line :

1 1y y m x x And , the equation of straight line if two points are given 1 1 2 1x , y & x ,x :

1 2 1

1 2 1

y y y y

x x x x

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Introduction

First: Geometric Applications

Notes


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L1

L2

L3

point of tangency

we want to find the equation of line passes through a point on a curve :

Then the equation of line here is called: The equation of tangent line on the curve

I ts Rule is:

Where

1 1dy

m Tan f ' x Slope of tangent line And x ,y is called the point of tangencydx

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1 1

x ,y

11 The equation of the normal line to the tangent is : y y x x

m

0 If the tangent is // tox axisdy

2 m Tan If makes a ve angle with the x axisdx

1If the tangent is to x axis

0

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Remarks

But what if

1 1y y m x x


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Example (1)

Consider the function f(x) = x - 6x + 9x , then Find the equation of the tangent line to the

Curve of f at x = 4.

Answer The equation of the tangent : y y m x x 1 1

Where 1x 4 and y 1 f(4) = (4)6(4) + 9(4) = 4

And m = f `(x) = 3x - 12x + 9At x = 4 : m = f `(4) = 3(4) - 12(4) + 9 = 9

Then Substitute: y4 = 9( x4) y4 = 9x36Finally the equation of the tangent is : y = 4x32

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Example(2)Find the points on the curve 2y x 4 x 3 , at which the tangent to this curve is :

(1) Parallel to the xaxis

(2) Parallel to the line : y= 2 x + 3

(3) Perpendicular to the line 6 y 3x 5

Answer

(1) Parallel to xAxis meansdy

m 0dx

2

dy

2x 4 2x 4 0 2x 4 x 2dx

Subsitute in the orignal : y 2 4 2 3 -1

Then the point that the tangent passes with is 2 ,-1

2

22 m is the slope of line y 2x 3 2x 6 x 3

1

Substitute in the original equation : y 3 4 3 3 0 Thus The point 3,0

2

3 1

3 m is the slope of 6 y 3x 5 Then m -26 2

dy2x 4 2 2x 2 x 1

dx

Subtitute in the equation : y 1 4 1 3 0 Thus the Point is 1,0

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2 2 2 2

y 2x -1 -2y x y 2x 3y 3x2 y x

x y

Then substitute in the equation of the curve :

x x x 3 0 x 3

Then x 3 Or x - 3

y 3 Or y - 3

Then the points on the curve are : 3 , 3 and - 3 , - 3

Example(3)

Find the points on the curve 2 2x x y y 3 0 , at which the tangent to this curve is parallel

To the line y = - x .

AnswerFor the liney = - x :The slope = -1 - - - (1)

For the Curve2 2x x y y 3 0 :

dy dyThe slope : 2x x y 2y 0

dx dx

dy dy y 2x2 y x y 2x 2

dx dx 2 y x

from (1) and (2) :

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Example(4)

Find The slope of the tangent to the curve of the function 2 2x 3xy 5 y 3 at x = 2 which

lies on it.

Answer

At 2 2x 2 2 3 2 y 5y 3

2 25y 6 y 4 3 0 5y 6 y 1 0

15y 1 y 1 0 y - , y -1

5

1The two points 2 , - & 2, 1 are the points of tangency

5

To find The slope here means to get the differentiation with Respect to xdy dy

2x 3x 3y 10 ydx dx

12 ,

5

2, 1

dy dy0 3x 10 y - 2x 3y

dx dx

- 2x 3ydy dy3x 10 y - 2x 3y

dx dx 3x 10 y

1So,at 2 , - :

5

1- 2 2 3 - 5dy

The slope -0.85-1dx

3 2 105

At 2 ,-1 :

- 2 2 3 -1dy -1 1the slope

dx 3 2 10 -1 -4

4

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Example(5)Find the equation of the tangent and of the normal to the curve 3 2y x 2x 5 at the point

x -1 which lies on itAnswer

1 1

3 2

1 1

2

2

x -1

the equation of the tangent line : y y m x x

At x -1 y -1 2 -1 5 2 Then the point of tangency is -1,2

dyAlso m 3x 4x

dx

dyThen The slope of the tangent at -1 , 2 : 3 -1 4 -1 7

dx

the equation o

f the tangent line : y 2 7 x 1 y 7x 9-1

And the slope 7 , Then the normal Slope is7

-1the equation of the normal line : y 2 x 1

7

1 1 1 13y - x 2 y x 7 7 y x 13 0

7 7 7 7

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Example(6)

Find the equation of the tangent to the curve2

tan xy at point x

1 tan x 3

Answer

2 2

1 1

2

1 1

2

x3

2tan x 1 Tan x 1Tan Tan2x y Tan2x

1 tan x 2 1 Tan x 2

The equation of the tangent line : y y m x x where

1 3 dy

x & y Tan2 - and m Sec 2x3 2 3 2 dx

dySo at x Slope Sec 2 4

3 dx 3

3y

2

4 3

4 x y 4x 6 6 y 24x 8 3 33 3 2

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1 1y y m x x

1 1x ,y

-2,0

2

1 1 1 1So to get y at x : y 8x 1

Example (7)

Find the two equations of the two tangents to the curve 2y 8x which passes through (-2,0)

and prove that the two tangents are perpendicular .

AnswerThere is a big di fference between :

Equation of a tangent of the curve at ( x , y) direct substitution in the ruleEquation of a tangent of the curve which passes through (x , y) find the tangency point firstSo in our problem, we dont have tangency points , so we must get them

Let the point of tangency between the tangent and the curve be 1 1x , y So the equation of the first tangent with the curve :

and the slope of the tangent on the curve :dy dy 4

2y 8dx dx y

at 1 1x ,y 1

dy 4

dx y

But the slope of the tangent line passing through (-2,0) & 1 1x , y is :2 1 1

2 1 1

y y ym

x x x 2

211 1

1 1

1 1 1 1

2

1 1

y 4So , both slopes are equal , then : y 4x 8 2x 2 y

Substitue 2 in 1 : 8x 4x 8 4x 8 x 2

Substitute in 2 : y 16 y 4

Then the two points of tangency are : 2,4 and 2,-4

dy 4 4at point 2,4 :

dx y

14

equation of the first tangent : y 4=1 x 2 y x 2 4 y x 2

dy 4 4at point 2, 4 : -1

dx y -4

equation of the first tangent : y 4 -1 x 2 y -x 2 4 y -x 2

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Example(8)

If y Tan x where x 0,2 , find the points on the curve of this function at which theTangent is parallel to the straight line 2 y 8 x 7 0

Answer

2

2

dy 8for 2y 8x 7 m 4

dx 2

dyfor y = Tan x Sec x

dx

The tangent to the curve is parallel to line

Then they have the same slope : Sec x 4

Sec x 2

Sec x 2 Or Sec x -2

1Cos x Or

2

1 o 1 o

o o o o o o

o o

o o

o o

1Cos x -

2

1 1x Cos 60 , x Cos - 120

2 2

Or x 360 60 300 , x 360 120 240

Then when x 60 y Tan60 3

And when x 120 y Tan120 - 3when x 240 y Tan240 3

Thus the points on t

o o o ohe curve are : 60 , 3 , 300 , - 3 , 120 , - 3 , 240 , 3

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Example (9)Prove that the point ( -1,3) lies on the curve 2 2x y 4 x 2 y 20 and then find the

equation of the tangent and the normal line to the curve at this point.

AnswerSubstitute The point (-1,3) on the curve 2 2x y 4 x 2 y , If its value was 20 as the R.H.S

Then the point lies on the curve .

So substitute by (-1,3) : 2 2

-1 3 4 -1 2 3 20 R.H.S

Then the point lies on the curve .

The Equation of tangent line : 1 1y y m x x

To get the slopedy

: differentiate with respect to xdx

dy dy dy dy2x 2y 4 2 0 2y 2 4 2xdx dx dx dx

2 2 xdy dy 4 2x2y 2 4 2x

dx dx 2y 2 2 y 1

dy 2 x dy 2 1 3, So at -1,3 :

dx y 1 dx 3 1 4

3 3 3at -1,3 : the equation of tangent lineis : y 3 x 1 y x 3

4 4 43 15

y x 4 4 y 3x 154 4

0

-4The equation of normal Whose slope m

3

-4y 3 x 1 3 3y 9 -4 x 1 3y -4x 4 9 3y 4x 5 0

3

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Example (10)

Find the equation of the tangent and the normal to the curve 2y x 2 x 3 at its points of

intersection with the y x 1 0

AnswerFind the point of intersection between the curve and the line .

2y x 2x 3 1 & y x 1 2

Substitute (2) in (1):

2 2x 1 x 2x 3 x x 2 0 x 2 x 1 0

x -2 , x 1

y -3 , y 0

the points of intersection between the curve and the line : -2,-3 & 1,0

Then the equation of tangent line of the curve at -2,-3 :

1 1

dyy y m x x Wher m 2x 2

dx

dySo at -2 ,-3 2 -2 2 -2

dx

y 3 -2 x 2 y -2x 4 3 y -2x 7

1 1

1

Also ,The equation of Normal line with m 2

1y 3 x 2 2 2y 6 x 2 2y x 4

2

The equation of the tangent line of the curve at 1,0 :

dyy y m x x Where m 2x 2 y 4x 4

dx

1The equation of normal line : y - x 1 4

4

4 y

- x 1 4 y x 1 0

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2 2

2 2

2

2

Touch each other means that they have a

y 2x x 1 1 & y x x 2

Substitute 1 and 2 : 2x x 1 x x

x 2x 1 0 x 1 x 1 0 x -1

substitute to get y : y -1 -1 2

both curves t

common point and tangent

2

2

ouch each other in one point -1,2

dy dyfor y 2x x 1: 4x 1 at -1,2 4 -1 1 -3

dx dx

dy dyfor y x x : 2x 1 at -1,2 2 -1 1 -3

dx dx

So , both curves touch at one point -1,2 and the slope of their tangents is the same -

1 1

3

Then they have a common tangent :

Its equation is : y y m x x y 2 -3 x 1 y -3x 3 2 y -3x 1

Curve 1

Curve 2

Common tangent Common point

Example (11)

Prove that the two curves 2 2y 2x 3x 8 and y x 3x 9 intersect orthogonally at x=1

Answer

To prove that the two curves intersect orthogonally :First: we have to prove that both curve intersect.

Second: prove that the product of their slopes 1

2

2

x 1

2

So for y 2x 3x 8 :

Substitute by x 1 in the equation : y 2 1 3 1 8 7

So the curve passes through 1,7

dy dySlope of the tangent : 4x 3 , So at x 1: 4 1 3 1

dx dx

for y x 3x 9 :

by substituting in the equatio

2

x 1

n by x 1 : y 1 3 1 9 7

So the curve passes through 1,7

dy dyThe slope of tangent : 2x 3 , So at x 1: 2 1 3 -1

dx dx

The two curves intersect at 1,7 and thier tan gents are perpendicular

Then the two curves

intersect orthogonally.

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Example (12)

Prove that : the two curves 2 2y 2x x 1 and y x x touch each other , and find the

equation of the common tangent at the point of tangency.

Answer


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Example (13)

If the two curves 2y x ax b and 3 2y x x x c touch each other at the point (-1,0),

Then find the numerical values of a , b , c , find also the equation of their common tangent at

this point.Answer

Both curves touch each other at a point , then this point satisfy the both curves , Also both

curves have a common tangent.

2

x 1

3 2

3 2

2

So for y x ax b :

at -1,0 : 0 1 a b b a -1 1

dythe slope of its tangent at -1,0 is : 2 x a

dx

dy-2 a 2

dx

for y x x x c

at -1,0 : 0 -1 -1 -1 c 0 -1 1 1 c c 1

dyAnd 3x 2x 1

dx

dyat x -1 3dx

2 1 4 3

from 2 and 3 : -2 a 4 a 6

Substitute in 1 : b 6 -1 b 5

The equation of the common tangent : y 4 x 1 y 4x 4

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Example (14)

If x 0 , ,Then find a point on a curve y Sin2x Cos x such that the tangent at it to thecurve makes an angle of measure 135 with the +ve direction of x-Axis , find also the equation

of this tangent.Answer

o

2 2

2 2

2

dythe slope of the tangent Tan Tan135 -1 1

dx

dyand 2Cos 2x Sin x 2

dx

from 1 & 2 : 2Cos 2x Sin x -1

Cos 2x 1 2Sin x 2 1 2sin x sin x -1

2 4 sin x sin x 1 0 -4 sin x sin x 3 0 -1

4 sin x sin x 3 0 4 sin x 3 si

1 o

o o o 1

n x 1 0

3 3sin x - x sin - -48 35' " refused" Or sin x 1

4 4

Or x 180 -48 35' 228 35' " refused" x sin 12

So x " agreed" substitute to get y y sin cos 02 2

Then the point is ,02

Then the equation of tangent line : y -1 x y -x2 2

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a ,b

2,-2

2 2let the point of tangency between the tangent and the curve be a , b

This means that a ,b lies on the curve a b 16 1

dyThen The slope of the tangent on the curve : 2x 2y 0

dx

dy dy x

x y 0dx dx y

Then the sl

2 1

2 1

2 2 2 2

dy aope of the tangent at a ,b : 2

dx b

y yAnd the slope of the tangent passes through 2 ,-2 and a ,b : m

x x

b 2m 3

a 2

b 2 afrom 2 and 3 : b 2b a 2a a b 2b 2aa 2 b

from 1 : 2a 2b 16 a b 8 a 8 b

2 2 2 2

4

Then substitute in 1 :

8 b b 16 64 16b b b 16 16b 48 b 3 Then a 5

dy 5Then the point of tangency is 5 ,3 and the slope is

dx 3

Then the equation of the tangent line of the curve is :

5 5 2y 3 x 5 y x

3 3

5

3 3y 5x 16 03

Example (15)

2 2Find the equation of the tangent to the circle x y 16 which passes through the point 2 ,-2

Answer

Again There is a big di fference between :Equation of a tangent of the curve at ( x , y) direct substitution in the ruleEquation of a tangent of the curve which passes through (x , y) find the tangency point first

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1x ,0 2x ,0A B

1

2

1 1 1 1

1 1

The curve intersects the x - axis at A , B , Then

Let x ,0 be the point of tangency between the tangent

and the curve ,Then at : x , 0 :

3 x 7x 4 0 3x 4 x 1 0

4Then x Or x 1

3

Then we have two ta

2

4ngents , their point of tangency are , 0 and 1 , 0

3

So for the curve : y 3x 7x 4

dyThe slope of the tangent is : 6 x 7

dx

4 dy 4So at , 0 : 6 7 1

3 dx 3

dy

and at 1 , 0 : 6 1 7 -1dxSo the product of b

oth slope of the two tangents is -1 Then they are Orthogonal

Example (16)2If the curve y 3x 7x 4 intersects the x - axis in two points A and B , Then prove that

the two tangents at A and B are orthogonal .

Answer

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EA

KBFD

M20 2018

C

EA

KBFD

M20 2018

C 0 18,

0 0,

15 20,

1515

E is the lowest point in the quadratic function Then it is the vertex of the parbola

So let F 0 , 0 So E 0 ,18 and A 15 ,20

Then we have to get the equation of parabola passes

through 15, 20 with the vertex 0 ,1

2

2

2

2

8

So y a x b c

220 a 15 18 225a 2 a

2252

Then the equation of the parabola is : y x 18225

Let us find Now the equation of the tangent line MA to the curve :

2The slope of the tangent to the curve y x

225

2

18 :

dy 4 dy 4 4x at 15 , 20 15

dx 225 dx 225 15

Then the equation of tangent line at 15, 20 :

4 4y 20 x 15 y x 1615 15

At x 0 Then y 16 FM 16 m

1Then the area of the trapeziod ABFM : 16 20 15 270 m

2To get BK : l

et K x ,0

dy -15Then the equation of the normal line AK to the curve at 15,20 and Slope is :

dx 4

-15 -15 225 -15 305y 20 x 15 y x 20 y x 4 y -15x 3054 4 4 4 4

305 61So put y 0 , Then 15x 305 x cm T

15 3

61 16

hen BK 15 cm3 3

A cable CA is suspended from two vertical poles AB and CD As shown in the figure ,eachof height 20 m and are 30 m apart from each other

If the cable has the shape of a quadratic function

and the point E is the lowest point on it , and it is

fixed by a normal wire AK to it such that :

K , B , F and D are collinear and EF 18 cm

Then find BK , and if the tangent to the cable meets

EF at M , Then find the

area of the trapezoid ABFM

Answer

For Excellent Pupils


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1x ,0 2x ,0

A B

1

2

1 1 1 1

1 1

The curve intersects the x - axis at A , B , Then

Let x ,0 be the point of tangency between the tangent

and the curve ,Then at : x , 0 :

3 x 7x 4 0 3x 4 x 1 0

4Then x Or x 1

3

Then we have two ta

2

4ngents , their point of tangency are , 0 and 1 , 0

3

So for the curve : y=3x 7x 4

dyThe slope of the tangent is : 6 x 7

dx

4 dy 4So at , 0 : 6 7 13 dx 3

dyand at 1 , 0 : 6 1 7 -1

dxSo the product of b

oth slope of the two tangents is -1 Then they are Orthogonal

General problemsExample (1)

2If the curve y=3x 7x 4 intersects the x - axis in two points A and B , Then prove that

the two tangents at A and B are orthogonal .

Answer

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Example (2)

Find the equation to the normal for the curve 2y x 3x 5 at each point of intersection2 2with the circle x 3x y 25 .

Answer

2 2 2

2

2 2

2 2

y x 3x 5 1 And x 3x y 25 2

From 1 : x 3x y 5

Substitute 3 in 2 : y 5 y 25 y y 30 0

Then y 6 y 5 0 y -6 Or y 5

To get x :

When y -6 :

Substitute in 3 : x 3x -11 x 3x 11 0 refused When y 5 :

Substitute in 3 : x

2

2

3x 0 x x 3 0

x 0 and x 3

Then the points of intersection are 0 ,5 and 3 , 5

dyFor the curve : y x 3x 5 The slope : 2x 3

dxdy 1

at 0 ,5 : 2 0 3 -3 The slope of normal

dx 3

Then the equation of the n

1ormal : y 5 x 3y 15 x 3y x 15 0

3dy 1

at 3 ,5 : 2 3 3 3 The slope of normal -dx 3

1Then the equation of the normal : y 5 - x 3 3y 15 -x 3 3y x 18 0

3

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Example (3)2 2Find the equation of the two tangents to the circle x y 5 Such that they are inclined to

the +ve x - axis by an angle whose tangent value equals 2

Answer

2 2

2 2 2 2 2 2

For the curve x y 5:

dy dyThe slope of its tangent : 2x 2y 0 2 x y 0

dx dxdy -x -x

Then and Tan 2 Then 2dx y y

Then x -2 y

Substitute in the original curve :

-2y y 5 4 y y 5 5y 5 y 1 y 1

So at y 1 x

-2

And y -1 x 2

The points of tangency are -2 , 1 and 2 , -1

At -2 ,1 And the slope is 2

The equation of the tangent here is y 1 2 x 2 y 2x 5

At 2 ,-1 And the slope is 2

The equation of the tangent here is y 1 2 x 2 y 2x 5

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Example (4)

Find the equation of the normal to the curve y xTan x Sin2x x at the pointx4

which

lies on the curve.

Answer2

1 1

2

The slope of the curve: y' x Sec x Tanx 2Cos 2x 1

At x : y Tan Sin 1 14 4 4 2 4 4 4

at ,1 : y' Sec Tan 2Cos 1 1 0 14 4 4 4 2 2 2

-2Slope of the normal at , 1

4

Equation of the normal is

-2 3: y 1 x y 2x

4 2

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Example(5)

Find the equation of the tangent and the normal to the curvex

y Tan at the point2

3

x2

which lies on the curve . Answer

:

1 1

1 1 2

2

2

Equation of the tangent : y y m x x

3 3 3x y Tan -1 Then the point of tangency is , -1

2 2x 2

dy 1 xThen the slope of tangent : m sec

dx 2 2

3 dy 1 3at x sec 12 dx 2 4

The equation of the tangent : y 1

3 3 3

1 x y x 1 y x 1 02 2 2

The equation of the normal with slope -1 :

3 3 3y 1 -1 x y -x 1 y x 1 0

2 2 2

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Example (6)

3 2If the curve y= ax +bx cx d , where a , b , c , d are Constants , passes through 1,0

3,-4 And 4 ,0 and if the tangent to it at the first point is parallel to the x - axis , Then

find a , b , c and d

Answer

2 2

1,0 , 3 ,-4 , 4 ,0 lie on the curve

At 1,0 a b c d 0 1

At 3 ,-4 27a 9b 3c d -4 2

At 4 ,0 64a 16b 4c d 0 3And the tangent at 1,0 are parallel to x - axis :

dy dy0 3ax 2bx c 3ax 2bx c 0

dx dxThen at 1,0 : 3a 2b

c 0 4

Subtract 2 1 : 26a 8b 2c -4 13a 4b c -2 5

Subtract 5 4 : 10a 2b -2 5a b -1 6

Subtract 3 2 : 37a 7b c 4 7

Subtract 7 5 : 24a 3b 6 8 a b 2 8

Subtract 8 6 : 3a 3 a 1 Substitute in 6 : b -6 Substitutein 4 : 3 12

c 0 c 9 Then d -4

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Example (7)

2 2

1 11 12 2 2 2

1 1

x y x x y yProve that the equation of the tangent to 1 at point x ,y is 1 where

a b a b

x ,y lies on the curve .

Answer

2 2 2 2

1 11 1 2 2 2 2

1 11 12 2 2 2

2 2

1 1 1 1 1 1

2 2 2 2 2 2

1 1

x y x yx ,y lies on the curve 1 1

a b a b

2 2 dy x y dyAlso the slope of the tangent to the curve : x y 0 2 0

a b dx a b dx

y dy x dy x y x b b x- -

b dx a dx a b a y a y

Then the equation of

2 2

2 21 1 11 1 1 1 1 12 2 2 2

1

the tangent line of the curve is :b x b x x y y

y y - x x y y y - x x x 1a y a a b

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Example (8)

2 22 2If the two curves x a y 18 and x a y 18 intersect orthogonally , then find a

Answer

Intersect orthogonally means that the two curves intersect with each other and the product of

their slopes is -1

2 22 2

2 2 2 2 2 2

2 2 2 2 2

2

x a y 18 1 & x a y 18 2

Subtract 2 1 : x a x a 0 x 2a x a x 2a x a 0

4a x 0 x 0

Sunstitute to get y : a y 18 y 18 a y 18 a

Then the point of intersections between the two curves are 0 , 18 a Or

2

2 2

2 2

0 , - 18 a

for the curve x a y 18 :

-2 x ady dy a xThe slope is : 2 x a 2y 0

dx dx 2 y y

for the curve x a y 18 :

-2 x a - a xdy dyThe slope is : 2 x a 2 y 0

dx dx 2y y

athe two curves intersect orthogonally :

2 2 2

2 2 2 2 2

x yy a x

y a x

At 0 , 18 a :18 a a 2a 18 a 9 a 3

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