Application of Derivatives NCERT Solutions
NCERT Exercise 6.3 Q:15-27
Question: (15) Find the equation of the tangent line to the curve`y= x^2 - 2x + 7` Which is (a)parallel to the line `2x -y+9=0`, (b) perpendicular to the line `5y-15x=13`
Solution:
Here ` y=x^2 - 2x +7`
`(dy)/(dx)`
` = d/ (dx) ( x^2 - 2x +7)`
` =2x - 2= 2(x-1)`
(a) Slope of the line `2x-y+9=0` is
=`(text{-coefficient of}\ x)/(text{coefficient of}\ y)`
`=-2/-1 =2`
It is given that tangent is parallel to the line
`:.2(x-1)=2`
`=> x-1 =1`
` =>x = 2`
When `x=2` then y`
` y=(2)^2-2xx2+7`
`=4-4+7=7`
` :.` Equation of tangent at `(2, 7)` is
` y-7=2(x-2)`
`=> y-7 = 2x -4`
` =>2x-y+3=0`
(b)Slope of the line `5y-15x=13 `is
=`(text{-coefficient of}\ x)/(text{coefficient of}\ y)`
` =(-(-15))/5 =3`
It is given that tangent is perpendicular to the line
`:. 2(x-1)xx3=-1`
` => 6x - 6 =-1 `
`=> x= 5/6`
When `x= 5/6` then `y=(5/6)^2 -2xx5/6 +7 `
` (25-60+252)/36 = (217)/36`
`:. ` Equation of tangent at `(5/6, (217)/36)` is
` y= (217)/36 `
` = 1/3 (x-5/6)`
`=> 36y- 217 = - 36/3 (x-5/6)`
`=> 36 y - 217 =- 12x + 10`
12x+ 36 y - 227 =0`
Question: (16) Show that the tangents to the curve `y=7x^3 + 11` at the points where ` x=-2 ` are parallel.
Solution:
<Here ` y=7x^3+11`
`(dy)/(dx) = d/ (dx) (7x^2 + 11)`
`21 x^2`
Slope of tangent at point `x= -2` is
`[(dy)/(dx)](x=2)`
` 21 xx(2)^2 =84`
Slope of tangent at point ` x=-2 ` is
`[(dy)/(dx)](x=-2)`
` 21 xx(2)^2 =84`
Now slope of tangent at ` x=2 = `slope of tan gent at ` x=- 2`
Thus tangents to the curve at points ` x=2` and `x=-2 ` are parallel.
Question: (17) Find the points on the curve ` y= x^3` at which the slope of the tangent is equal to the ` y- coordinate of the point.
Solution:
Here ` y=x^3`
` (dy)/(dx)= d/(dx)(x^3)`
` 3x^2`
Let `P(x_1, y_1)` be any point on the curve ` y=x^3`
`:. y_1 = x_1^3`
Now `[ (dy)/(dx)](x=x_1) = 3x_1^2`
It is given that slope of tangent is equal to `y-` coordinate of the point
`:. y_1 =3x_1^2`
From (i)and (ii) we have
`x_1^3=3x_1^2`
`=> x_1^2- 3x_1 ^2 =0 `
`=> x_1^2(x_1-3)=0 `
`=> x_1=0 ` or `x_1=3`
when `x_1 = 0` then
`y_1= (0)^3 =0`
When `x_1 = 3` then `y_1= (3)^3 =27`
Thus required points are `(0, 0)` and `(3, 27)`
Question: (18) for the curve ` y= 4x^3 - 2x^5`, find all the points at which the tangent passes through the origin.
Solution:
Here `y=4x^3-2x^5`
`(dy)/(dx)= d/(dx)[4x^3- 2x^5]`
` 12x^2-10x^4`
Let `p(x_1,y_1)` be any point on the curve `y= 4x^3- 2x^5`
`:. y_1= 4x_1^3-2x_1^5`
Slope of tangent at point `p(x_1,y_1)` is
`[(dy)/(dx)]_(x=x_1)`
`= 12x_1^2 -10 x_1^4`
Equation of tangent at point `(x_1,y_1)` is
`y-y_1 = (12x_1^2-10 x_1^4)(x-x_1)`
Since the tangent passes through origin
`:.0-y_1`
`=(12x_1^2 - 10x_1^4)(0-x_1)`
`=> y_1 `
`= x_1(12x_1^2-10 x_1^4)`
`= 12x_1^3 10x_1^5`
From (i)and (ii) ,we have
`4x_1^3-2x_1^5`
` =12x_1^3 - 10x_1^5`
`=> -2x_1^5+10x_1^5`
`=12x_1^3-4x_1^3`
`=> 8x_1^5`
`=8x_1^3`
`=>8x_1^5- 8x_1^3=0`
`8x_1^3(x_1^2-1)=0`
`=> 8x_1^3(x_1+1)(x_1-1)=0`
` =>x_1=0 ` or `x_1=-1` or ` x_1=1`
When `x_1=0` then `y_1`
`=4xx(0)^3-2(0)^5=0`
When `x_1 =-1` then ` y_1 =4xx(-1)^3-2(-1)^5`
`=-4+2=-2`
When `x_1=1 ` then `y_1=4xx(1)^3-2(1)^5`
` 4-2 =2`
Thus the required points are `(0,0), (-1, -2)` and `(1,2)`
Question: (19) Find the points on the curve `x^2+y^2-2x-3=0` at which the tangents are parallel to the `x- `axis.
Solution:
The equation of given curve is `x^2+y^2-2x-3=0`
Differentiating both sides w.r.t. `x`, we have
`d/(dx)(x^2+y^2-2x-3)`
` = d/(dx)(0)`
` Differentiating both sides w.r.t. `x` ,we have
` d/(dx)(x^2+y^2-2x-3)=d/(dx)(0)`
` 2x+2y (dy)/(dx)-2=0`
`=> (dy)/(dx)`
`=(2-2x)/2y =(1-x)/y`
Since the tangent is parallel to `x-`axis
` :.(dy)(dx)=0`
Now `(1-x)/y =0 `
=> 1-x=0 => x=1`
When `x=1` then `(1)^2 + y^2-2xx1 -3=0`
`=>y^2 =4 => y=+-2`
Thus required points are `(1+-2)`
Question:(20) Find the equation of the normal at the point `(am^2,am^2)` for the curve `ay^2=x^3`
Solution:
Here `ay^2=x^3`
Differentiating both sides w.r.t. `x` ,we have
`d/(dx) (ay^2)`
` = d/(dx)(x^3)`
`2ay(dy)/(dx) `
` 3x^2=>(dy)/(dx)=(3x^2)/2ay`
`[(dy)/(dx)]`at `(am^2, am^3)=`
`(3(am^2)^2)/(2a(am^3)`
`=(3a^2m^4)/(2a^am^3)`
`=3/2 m`
Equation of normal at point `(am^2, am^3) ` is
`y-am^3`
` =-1/(3m)/2 (x-am^2)`
`=>y-am^3=-2/(3m)(x-am^2)`
`3my-3am^4`
`=-2x+2am^2`
` => 2x + 3my-3am^4-2am^2=0`
`=>2x+3my-am^2(3m^2+2)=0`
`=> 2x+3my- am^2(3m^2+2)=0`
`Question:(21) Find the equation of the normals to the curve ` y=x^2+2x+6` Which are parallel to the line` x+14y+4=0`
Solution:
Here `y=x^3+2x+6`
`(dy)/(dx)=d/(dx)(x^3+2x+6)`
`=3x^2+2`
Slope of normal`=-1/((dy)/(dx))`
`=-1/(3x^2+2)`
Slope of the line `x+14y+4=0` is
=`(text{-coefficient of}\ x)/(text{coefficient of}\ y)`
`=- 1/14`
Since normal is parallel to the line
`:. -1/(3x^2+2)`
` =-1/14`
` => 3x^2+2 `
` 14 => 3x^2=12`
` =>x^2=4=>x=+-2`
When `x=2` then `y=(2)^3+ 2xx2+6`
` =8+4+6=18`
When `x=-2` then `y=(-2)^3+2xx -2+6`
`=-8-4+6=-6`
`:.` Equation of normal at point `(2,18)` is
`y-18=-1/ 14 (x-2)`
`=> 14y - 252 =-x+2`
`=> x+14y-254=0`
Equation of normal at point `(-2,-6)` is
`y+6 =-1/14 (x+2)`
`=> 14y+84=-x-2`
`=>x+14y+86=0`
Question: (22)Find the equations of the tangent and normal to the parabola `y^2 = 4ax` at the point `(at^2, 2at)`
Solution:
Here ` y^2=4ax `
Differentiating both sides w.r.t. `x` , we have
`2y (dy)/(dx) =4a `
`=> (dy)/(dx) `
`(4a)/(2y) =(2a)/(y)`
` [(dy)/(dx)] at (at^2, 2at)`
` = 2a/2at=1/t`
Equation of tangent at point `(at^2,2at)` is
` y- 2at`
` =1/t (x-at^2)`
`=>ty-2at^2=x-at^2`
`=> x-ty+at^2 =0`
Equation of normal at point `( at^2,2at)`is
` - 2at`
` =- 1/(1//t)(x-at^2)`
`=>y-2at =-t(x-at^2)`
` y-2at =-tx+at^3`
`=>tx+y-2at - at^3=0`
`=>tx+y - at(2+t^2)=0`
Question: (23) Prove that the curves `x=y^2` and ` xy=k` cut at right angles if `8k^2 =1`
Solution:
Here `x= y^2`
and `xy=k`
Putting value of `x` from (i)in (ii), we have
`y^2.y = k`
`y^3 =k`
` =>y=(k)^(1/3)`
Putting value of `y` in (i), we have
`x=[(k)^(1/3)]^2`
` =(k)^2/3`
`:.` point of intersection of two curves is `k^(2/3), k^(1/3)`
Differentiating both sides w.r.t. `x`, we have
`d/(dx)(x)= d/(dx)(y^2)`
`=2y (dy)/(dx) =1/(2y)`
`:.` Slope of tangent at `k^(2/3), k^(1/3)= 1/ 2k^(1/3)`
Differentiating both sides w.r.t. `x`, we have
`d/(dx)(x)= d/(dx)(y^2)`
`=2y (dy)/(dx)`
` => (dy)/(dx) 1/(2y)`
`:.` Slope of tangent at `k^(2/3), k^(1/3)= 1/ 2k^(1/3)`
Differentiating (ii)w.r.t `x`, we have
`d/(dx)(xy)=d/(dx)(k)`
` x (dy)/(dx) + y.1=0`
`=> (dy)/(dx) =-y/x`
:. slope of tangent at `k^(2/3), k^(1/3)= (-k^(1/3))/(k^ (2/3))`
`=-1/(k^(1/3))`
Sinnce the two curved intersect at right angle
`:. 1/(2k^(1/3)) xx - 1/(k^(1/3))=-1`
` => (-1)/(2k^(2/3)) =-1`
` =>2k^(2/3)=1`
cubing both sides, we have
`2k^(2/3)^3`
`(1)^3`
`=>8k^2=1`
Question: (24) Find the equation of the tangent ang normal to the hyperbola `(x^2)/(a^2)-(y^2)/(b)^2=1` at the point`(x_0,y_0)`
Solution:
Here equation of the given curve is `(x^2)/(a^2)-(y^2)/(b^2) =1`
Differentiating both sides w.r.t. `x` , we have
` d/(dx)[(x^2)/(a^2)- (y^2)/(b^2)]= d/(dx)(i)`
`(2x)/(a^2)-(2y)/(b^2)(dy)/(dx)=0`
`=> (dy)/(dx)`
`=-(2x)/(a^2)xx (b^2)/(-2y)`
`=(b^2x)/(a^2y)`
`(dy)/(dx) at (x_0, y_0)`
`=(b^x_0)/(a^y_0)`
`:. ` Equation of tangent at `(x_0, y_0)` is
`y- y_0 = (b^2 x_0)/(a^2 y_0) (x- x_0)`
`=> (yy_0)/(b^2)-(y_0^2)/(b^2)`
` = (x x_0)/(a^2)-(x^2_0)/(a^2)`
` => (x x_0)/(a^2)-(yy_0)/(b^2)`
` (x_0^2)/(a^2)-(yy_0)/(b^2)= (x_0^2)/(a^2)-(y_0^2)/(b^2)`
`=>(x x_0)/(a^2) -(yy_0)/(b^2)=1`
`[because (x_0^2)/(a^2) - (y 0^2)/(b^2)=1]`
Equation of normal at `(x_0, y_0)` is
` y-y_0 = 1/(b^2x_0 // a^2 y_0)(x-x_0)`
`=> y-y _0 =-(a^2 y_0)/(d^2 x_0)(x-x_0)`
` =>b^2 x_0 y - b^2x 0 y `=
` =-a^2xy_0 + a^2x _0y_0`
` =>a^2xy_0 + b^2 x_0 y`
` = x_0y_0(a^2+b^2x)`
` =>(a^2xy_0)/(x_0y_0)+(b^2x_0y)/(x_0y_0)=a^2+b^2`
`=> (a^2x)/(x_0)+ (b^2y)/(y0) =a^2 + b^2`
Question: (25) Find the equation of the tangent to the curve ` y = squrt(3x-2)` which is parallel to the line `4x-2y+5=0`
Solution:
Here `y= sqrt(3x-2)`
`(dy)/(dx)=d/(dx) (sqrt(3x-2))`
=`1/(2sqrt(3x-2)\\ xx d/ (dx)(3x -2)`
`=3/(2sqrt(3x-2))`
Slope of tangent `=3/(2sqrt(3x-2))`
Slope of the line ` 4x-2y+5=0` is
=`(text{-coefficient of}\ x)/(text{coefficient of}\ y)`
`= -4/-2 =2`
Since tangent is parellel to the line
`:. 3/(2sqrt(3x-2))`
2=> 3= 4sqrt(3x-2)`
`=>9=16(3x-2)`
`9= 48x -32`
`=>48x=41`
`=> x=41/48`
When `x=41/48` then
` y=sqrt(3xx 41/48 -2)`
`= sqrt((123-96)/48)`
`=sqrt(27/48)`
`=sqrt(9/16)`
`= 3/4`
`:.` Equationof tangent at `((41)/(48), 3/4)` is
`y-3/4=2(x- 41/48)`
`=> y-3/4 =2x- 41/24`
`=> 2x-y- 41/24 +3/4 =0 `
`=> 2x-y - 23/24 =0 `
`=> 48x -24 y -23 =0`
Choose the correct answer in Exercise `26` and `27`
Question: (26) The slope of the normal to the curve `y=2x^2+3 sin x at x=0` is
(a)`3`
(b)`1/3`
(c)`-3`
(d)`- 1/3`
Solution:
Here `y= 22x^2+33sin x`
` (dy)(dx) = d/ (dx (2x^2+3 sin x)`
`= 4x+3 cos x`
` [(dy)/(dx)]_(x-0) = 4xx 0 + 3 cos 0=0+3xx1=3`
`;.` Slope of normal
`=- 1/[(dy/dx)]_(x-0)= -1/3`
Thus answer is (d)`
Question: (27) The lines `y=x+1` is tangent to the curve `y^2 = 4x ` at the point
(a)`(1, 2)`
(b)`(2,1)`
(c)`(1,-2)`
(d)`(-1,2)`
Solution:
Here `y= 4x`
`:. d/(dx(y^2) =d/(dx (4x)`
`2y(dY)/(dx)`
`=4 => (dy)/(dx)`
`=4/(2y) =2/y`
Slope of the line ` y=x+1 is 1`
`:. 2/y=1 => y=2`
When `y=2` then ` 2=x+1 => x=1`
Thus required point is `(1,2)`
Thus answer is (a)
Reference: