Question

14.Find the equations of the tangent and normal to the given curves at the indicatedpointsG) y-- 6r3 + 13*? - 10x + 5 at (0, 5)(i) y-rt-6r*+ rtd.3(ii) y-x3 at (, 1)(iv) y x2 at (0,0)(v) x- cost, y - sint at t-

Answer 1

The equation of the given curve is,

y= x 4 −6 x 3 +13 x 2 −10x+5

Slope of the normal to the curve is given as,

Slope of the normal= −1 Slope of the tangent (1)

Slope of tangent to a curve y is given by,

Slope=( dy dx )

Hence, the tangent to the curve is given as,

d( x 4 −6 x 3 +13 x 2 −10x+5 ) dx =4 x 3 −18 x 2 +26x−10 (2)

The slope of the tangent at point ( 0,5 ) is,

=4 ( 0 ) 3 −18 ( 0 ) 2 +26( 0 )−10 =−10

The equation of the tangent at ( x 1 , y 1 ) with slope m is given as,

y− y 1 =m( x− x 1 )(3)

Substitute ( 0,5 ) for ( x 1 , y 1 ) and −10 for m.

y−5=−10( x−0 ) y−5=−10x 10x+y=5

Thus, the slope of the normal from equation (1) is given by,

Slope of the normal= −1 −10 = 1 10

Equation of the normal at ( 0,5 ) is given by,

y−5= 1 10 ( x−0 ) x−10y+50=0

The slope of the tangent at point ( 1,3 ) from equation (2) is,

=4 ( 1 ) 3 −18 ( 1 ) 2 +26( 1 )−10 =2

Substitute ( 1,3 ) for ( x 1 , y 1 ) and 2 for m in equation (3).

y−1=2( x−3 ) y−1=2x−6 y=2x−5

Thus, the slope of the normal from equation (1) is given by,

Slope of the normal= −1 2 = −1 2

Equation of the normal at ( 1,3 ) is given by,

y−3= −1 2 ( x−1 ) x+2y−7=0

The equation of the given curve is,

y= x 3

Slope of tangent to a curve y is given by,

Slope=( dy dx )

Hence, the tangent to the curve is given as,

d( x 3 ) dx =3 x 2

The slope of the tangent at point ( 1,1 ) is,

=3 ( 1 ) 2 =3

Substitute ( 1,1 ) for ( x 1 , y 1 ) and 3 for m in equation (3).

y−1=3( x−1 ) y=3x−2

Thus, the slope of the normal from equation (1) is given by,

Slope of the normal= −1 3 = −1 3

Thus, equation of normal is given by,

y−1= −1 3 ( x−1 ) x+3y−4=0

The equation of the given curve is,

y= x 2

Slope of tangent to a curve y is given by,

slope=( dy dx )

Hence, the tangent to the curve is given as,

= d( x 2 ) dx =2x

The slope of the tangent at point ( 0,0 ) is,

2( 0 )=0

Substitute ( 0,0 ) for ( x 1 , y 1 ) and 0 for m in equation (3).

y−0=0( x−0 ) y=0

Thus, the slope of the normal from equation (1) is given by,

Slope of the normal= −1 0

As ( −1 0 ) is not defined, the slope of the normal must not be defined and the equation of the normal at ( 0,0 ) is x=0.

The equation of the curve is defined as,

x=cost y=sint (4)

The equation of tangent is defined as,

Slope= dy dx

As the curve is in parametric form, hence to equate slope we must partially derivate x and y with respect to t such that,

dy dx = ( dy dt ) ( dx dt ) (5)

The derivative of x with respect to t is,

dx dt = d( cost ) dt =−sint

The derivative of y with respect to t is,

dy dt = d( sint ) dt =cost

Substitute cost for dy dt and −sint for dx dt in equation (5).

dy dx = cost −sint =−cott

The slope of tangent at t= π 4 is,

−cot π 4 =−1

Substitute −1 for t in equation (4).

x=cos π 4 = 1 2 y=sin π 4 = 1 2

Thus, substitute ( 1 2 , 1 2 ) for ( x 1 , y 1 ) and −1 for m in equation (3).

y− 1 2 =−1( x− 1 2 ) x+y− 2 =0

Thus, the slope of the normal will be,

= −1 −1 =1

Therefore, equation of normal at ( 1 2 , 1 2 ) is,

y− 1 2 =1( x− 1 2 ) x=y