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APPLICATIONS OF DERIVATIVES
TANGENT AND NORMAL
mathews.suman@gmail.com
P
mathews.suman@gmail.com
Given a curve y = f(x), slope of the tangent to the curve at ),( 11 yx
is
),( 11
)( yx
dx
dy
Slope of the normal to the curve at ),( 11 yx is
)( 1,1
)(
1
yx
dx
dy
Equation of the tangent : )()( 1)(1 1,1
xx
dx
dy
yy yx 
Equation of the normal:
)(
)(
1
1
),(
1
1
xx
dx
dy
yy
yx

mathews.suman@gmail.com
Find the point on the curve
123 2
 xxy at which the slope
of the tangent is 4
26  x
dx
dy
6x -2 = 4
6x = 6
x = 1
y = 2
Required point =( 1,2)
mathews.suman@gmail.com
Find the points on the curve 1322
 yx the tangent at each
one of which is parallel to the line 2x + 3 y = 7
3
2
3
7
3
2
723
732






m
xy
xy
yx
y
x
dx
dy
x
dx
dy
y
dx
dy
yx
dx
dy
yx
yx





0
022
1322
3
2


y
x
yx
3
2

mathews.suman@gmail.com
𝑥2
+ 𝑦2
= 13
4
9
𝑦2 + 𝑦2 = 13
13𝑦2
9
= 13
y=±3
x=
2
3
3 = 2 One point of intersection = (2,3)
𝑥 =
2
3
−3 = −2 Other point of intersection = (-2,3)
mathews.suman@gmail.com
The acute angle between the 2 curves is the angle between the tangents
to the 2 curves at their point of intersection.
21
21
1
tan
mm
mm



121 mm The curves cut each other orthogonally
21 mm  The curves touch each other
mathews.suman@gmail.com
Find the condition that the curves
2
2 yx 
and 2xy =k intersect orthogonally
Solving both equations simultaneously
Point of intersection = ),
2
( 3/1
3/2
k
k
ky 3
3/1
ky 
2
2
y
x 
mathews.suman@gmail.com
For the 1st curve 22 
dx
dy
y
13/1
11
m
kydx
dy

For the 2nd curve, kxy 2
x
k
y
2

2
2x
k
dx
dy 

3/4
2
4
k
k

23/1
2
m
k



mathews.suman@gmail.com
121 mm
1
2
3/13/1


kk
23/2
k
82
k
mathews.suman@gmail.com
Prove that the curves xy = 4 and 822
 yx touch each other
We first find the point of intersection of the 2 curves
y
x
4
 8
16 2
2
 y
y
24
816 yy  0168 24
 yy
0)4( 22
y 42
y 2y
Point of intersection = (2,2) (-2,-2)
When y = 2, x=2 when y = -2, x= -2
mathews.suman@gmail.com
x
y
4

1m at (2,2) is -1
22
8 xy 
x
dx
dy
y 22 
y
x
dx
dy

  1)2,2(2 m
Since 21 mm  the curves touch each other
2
4
xdx
dy

mathews.suman@gmail.com
It can also be shown that at (-2,-2), the curves touch each other.

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Tangent and normal

  • 1. APPLICATIONS OF DERIVATIVES TANGENT AND NORMAL mathews.suman@gmail.com
  • 3. Given a curve y = f(x), slope of the tangent to the curve at ),( 11 yx is ),( 11 )( yx dx dy Slope of the normal to the curve at ),( 11 yx is )( 1,1 )( 1 yx dx dy Equation of the tangent : )()( 1)(1 1,1 xx dx dy yy yx  Equation of the normal: )( )( 1 1 ),( 1 1 xx dx dy yy yx  mathews.suman@gmail.com
  • 4. Find the point on the curve 123 2  xxy at which the slope of the tangent is 4 26  x dx dy 6x -2 = 4 6x = 6 x = 1 y = 2 Required point =( 1,2) mathews.suman@gmail.com
  • 5. Find the points on the curve 1322  yx the tangent at each one of which is parallel to the line 2x + 3 y = 7 3 2 3 7 3 2 723 732       m xy xy yx y x dx dy x dx dy y dx dy yx dx dy yx yx      0 022 1322 3 2   y x yx 3 2  mathews.suman@gmail.com
  • 6. 𝑥2 + 𝑦2 = 13 4 9 𝑦2 + 𝑦2 = 13 13𝑦2 9 = 13 y=±3 x= 2 3 3 = 2 One point of intersection = (2,3) 𝑥 = 2 3 −3 = −2 Other point of intersection = (-2,3) mathews.suman@gmail.com
  • 7. The acute angle between the 2 curves is the angle between the tangents to the 2 curves at their point of intersection. 21 21 1 tan mm mm    121 mm The curves cut each other orthogonally 21 mm  The curves touch each other mathews.suman@gmail.com
  • 8. Find the condition that the curves 2 2 yx  and 2xy =k intersect orthogonally Solving both equations simultaneously Point of intersection = ), 2 ( 3/1 3/2 k k ky 3 3/1 ky  2 2 y x  mathews.suman@gmail.com
  • 9. For the 1st curve 22  dx dy y 13/1 11 m kydx dy  For the 2nd curve, kxy 2 x k y 2  2 2x k dx dy   3/4 2 4 k k  23/1 2 m k    mathews.suman@gmail.com
  • 11. Prove that the curves xy = 4 and 822  yx touch each other We first find the point of intersection of the 2 curves y x 4  8 16 2 2  y y 24 816 yy  0168 24  yy 0)4( 22 y 42 y 2y Point of intersection = (2,2) (-2,-2) When y = 2, x=2 when y = -2, x= -2 mathews.suman@gmail.com
  • 12. x y 4  1m at (2,2) is -1 22 8 xy  x dx dy y 22  y x dx dy    1)2,2(2 m Since 21 mm  the curves touch each other 2 4 xdx dy  mathews.suman@gmail.com It can also be shown that at (-2,-2), the curves touch each other.