EQUATION OF TANGENT AND NORMAL TO THE CURVE AT THE GIVEN POINT

Find the equation of the tangent and normal of the following curves

(i)  y  =  x2-4x-5, at x = -2

(ii)  y  =  x-sin x cos x, at x = π/2

(iii)  y  =  2sin23x, at x = π/6

(iv)  y  =  (1+sin x)/cos x, at x  =  π/4

Question 1 :

y  =  x2-4x-5, at x = -2

When x  =  -2, then

y  =  (-2)2-4(-2)-5

   = 4+8-5

y  =  7

dy/dx  =  2x-4

dy/dx  =  2(-2)-4

dy/dx  =  -8

So, the required point is (-2, 7)

Equation of tangent :

(y-y1)  =  m(x-x1)

(y-7)  =  -8(x+2)

 y-7  =  -8(x+2)

 y-7  =   -8x-16

8x+y-7+16  =  0

8x+y+9  =  0

Equation of normal :

(y-y1)  =  (-1/m) (x-x1)

(y-7)  =  (1/8) (x+2)

 8(y-7)  =  1(x+2)

8y-56  =  x+2

x-8y+58  =  0

Question 2 :

y  =  x-sin x cos x, at x = π/2

Solution :

y  =  x-sin x cos x when x = π/2.

y =  π/2-sin π/2cos π/2

=  π/2 - (1)(0)

=  π/2 

dy/dx  =  1-[sin x (- sin x)+cos x (cos x)]

=  1-[- sin2x + cos2x]

=  1-[cos2x-sin2x]

=  1-cos 2x

Slope of the tangent at x  =  π/2

=  1-cos2(π/2)

=  1-cos π

=  2

So, the required point is (π/2, π/2)

Equation of tangent :

(y-y1)  =  m(x-x1)

[y-(π/2)]  =  2 [x-(π/2)]

[y-(π/2)]  =  2x- π

2x-y-π+(π/2)  =  0

2x-y-(π/2)  =  0

Equation of normal :

(y-y1)  =  (-1/m) (x-x1)

[y-(π/2)]  =  (-1/2) [x-(π/2)]

2[y-(π/2)]  =  -1[x-(π/2)]

2y-π   =  -x+(π/2)

x+2y-π-(π/2)  =  0

x+2y-(3π/2)  =  0

Question 3 :

y  =  2sin23x, at x = π/6

Solution :

y  =  2sin2 3x

dy/dx  =  2(2sin3x) (cos3x) 3

=  6(2sin3x cos3x)  

=  6sin 2(3x)

=  6sin6x

Slope at x  =  π/6

=  6 sin 6 (π/6)

=  6sin π

= 6(0)

=  0

y  =  2sin23x

=  2sin23(π/6)

=  2sin2(π/2)

=  2

So, the required point is (π/6,2)

Equation of tangent :

(y-y1)  =  m(x-x1)

(y-2)  =  0[x-(π/6)]

 y-2  =  0

Equation of normal :

(y-y1)  =  (-1/m) (x-x1)

(y-y1) = (-1/0) (x-x1)

0(y-2)  =  -1[x-(π/6)]

0  =  -1[x-(π/6)]

x - (π/6) = 0

Question 4 :

y  =  (1+sin x)/cos x, at x  =  π/4

Solution :

y  =  (1+sin x)/cos x

dy/dx  =  [cos x (cos x) - (1 + sin x) (-sin x)]/cos² x

=  [cos2x+sinx+sin2x]/cos2x

=  (1+sinx)/cos2x

Slope at x  =  π/4

y  =  (1+sinπ/4)/cos2π/4

=  (1+(1/√2))/(1/√2)2

=  [(√2+1)/√2]/(1/2)

=  [(√2+1)/√2]  (2/1)

=  (√2+1)√2

y  =  2+2√2

y  =  (1+sin x)/cos x

=  (1+sinπ/4)/cosπ/4

=  [1+(1/√2)]/(1/√2)

=  [(√2+1)/√2]/(√2/1)

=  (√2+1)

So, the required point is (π/4, (√2+1))

Equation of tangent :

(y-y1)  =  m(x-x1)

[y-(√2+1)]  =  (2+2√2) [x-(π/4)]

Equation of normal :

(y-y1)  =  (-1/m) (x-x1)

[y-(√2+1)]  =  [-1/(2+2√2)] [x-(π/4)]

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More