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Calculus
shahbaz ahmed
January 2024

Introduction

Reference to the figure
Triangle OEC is a right angle triangle with m̸ OEC = 90◦
...

Also m̸ EOC = α
m̸ OCB = α
m̸ BCD = 90 − α

1

m̸ BDC = α
m̸ COD = β
Also triangle OAD is a right angle triangle with m̸ OAD = 90◦
m̸ AOD = α + β
Now in the right angle triangle OAD
sin (α + β) =

AD
OD

sin (α + β) =

AB+BD
OD

Since AB = EC
sin (α + β) =

EC+BD
OD

sin (α + β) =

EC
OD

sin (α + β) =

EC OC
OC OD

+

BD
OD

+

BD DC
DC OD

2

sin (α + β) = sin α cos β + cos α sin β
...
eq (2)
Subtracting equation (2) from equation (1)
sin (α + β) − sin (α − β) = sin α cos β + cos α sin β − [sin α cos β − cos α sin β]
sin (α + β) − sin (α − β) = sin α cos β + cos α sin β − sin α cos β + cos α sin β
sin (α + β) − sin (α − β) = 2 cos α sin β
...
eq (4)
Putting α + β = x + δx,

α − β = x in equation (3)

3

sin (x + δx) − sin x = 2 cos x sin δx
2
...

(a) Show that the equation of the tangent to C at the point P is y = 1 − 2x
...
Solution
y = 9 − 4x − x8 , x > 0
at x=2
y = 9 − 4(2) −

8
2

= 9 − 8 − 4 = −3, x > 0

S0 (2, −3) is the point P on the curve
Taking derivatives with respect to x
dy
dx

=

8
d(9−4x− x
)
dx

dy
dx

=

d(9)
dx

dy
dx

= 0 − 4 d(x)
dx − 8

dy
dx

= 0 − 4 − 8 d(xdx

dy
dx

= 0 − 4 − 8(−1x−2 )

−

d(4x)
dx

−

8
d( x
)
dx

1
d( x
)
dx

−1

)

7

dy
dx

= 0 − 4 + 8(1x−2 )

dy
dx

= 0 − 4 + 8( x12 )

dy
dx

= −4 + ( x82 )

Putting x=2
dy
dx

= −4 + ( 282 ) = −2 = m

Where m is the slope of the curve at the point (2, −3) = (x1 , y1 )
Using the formula for slope m and one point (x1 , y1 )form of equation of line
y − y1 = m(x − x1 )
=⇒
y − (−3) = −2(x − 2)
y + 3 = −2x + 4
y = −2x + 4 − 3
y = −2x + 1 = 1 − 2x
(b) Equation of the normal to C at the point P
...

Let slope of normal to C at the point P be m1
Since normal and tangents are perpendicular to each other
mm1 =-1
Putting m = −2
−2m1 = −1
m1 =

1
2

Equation of line with slope m1 −

1
2

passing through the point (2, −3) = (x1 , y1 ) will be

y − (−3) = 21 (x − 2)
y + 3 = 12 (x − 2)
2(y + 3) = (x − 2)
2y + 6 = x − 2
x − 2y − 8 = 0
Putting y = 0 in the equation of tangent

8

y = −2x + 1 = 1 − 2x
1 − 2x = 0
x=

1
2

So Co-ordinates of the point A are
( 12
...
y1 ), B(8, 0) = (x2 , y2 ), P (2, −3 = (x3 , y3 )



x1 y1 1




1 
= 2 x2 y2 1




x3 y3 1
Putting A( 12 , 0) = (x1

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