applications on derivatives
TRANSCRIPT
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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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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
--------------------------------------------------------------------------------------------------------------------
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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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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Calculus Chapter (3) Appl ications on Deri vatives-2-
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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Calculus Chapter (3) Appl ications on Deri vatives-22-
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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Calculus Chapter (3) Appl ications on Deri vatives-011-
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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Calculus Chapter (3) Appl ications on Deri vatives-010-
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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Calculus Chapter (3) Appl ications on Deri vatives-019-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-01-
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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Calculus Chapter (3) Appl ications on Deri vatives-012-
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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Calculus Chapter (3) Appl ications on Deri vatives-001-
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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7/29/2019 Applications on Derivatives
20/21
Facebook page : https://www.facebook.com/MrSherifYehiaAlMaraghy
Mr . Sher if Yehia Al Maraghy https://twitter.com/Mr_Sheri f_yehia
0100998883601009988826 Email : Rooshery@hotmai l .com
Calculus Chapter (3) Appl ications on Deri vatives-000-
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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Facebook page : https://www.facebook.com/MrSherifYehiaAlMaraghy
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0100998883601009988826 Email : Rooshery@hotmai l .com
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