Application of Derivatives NCERT Solutions
NCERT Exercise 6.3
Question: (1)Find the slope of the tangent to the curve `y=3x^4 - 4x at =4`
Solution:
Here ` y= 3x^2 -4x`
`(dy)/dx = d/(dx) [3x^2 -4x]`
`= 12x^2 - 4`
`[(dy)/(dx)]_x=4 `
`=12xx(4)^3- 4`
` = 768 - 4 = 764`
Thus slope of tangent at `x=4` is `764`
Question: (2) Find the slope of the tangent to the curve `y= (x-1)/(x-2), x!= 2` at `x = 10`
Solution:
Here `y= (x-1)/(x-2)`
`(dy)/(dx) = d/ (dx)[(x-1)/(x-2)]`
`= ((x-2)d/(dx)(x-1)-(x-1)d/(dx) (x-2))/((x-2)^2)`
`= (x-2-x+1)/((x-2)^2)`
`=(-1)/((x-2)^2)`
` [(dy)/(dx)]_(x=10)`
` - 1/((10-2)^2)`
`= 1/(64)`
Thus slope of tangent at `x= 10 `is `1/(64)`
Question: (3) Find the slope of the tangent to curve ` y= x^2 -x+1` at the point whose `x` coordinate is `2`
Solution:
Here `y= x^2 -x -x+1`
`(dy)/(dx)= d/(dx)(x^3 - x+ 1)`
`= 3x^2 -1`
`[dy/dx]_(x=2)`
`= 3xx (2)^2-1 = 12 -1 =11`
Thus slope of tangent at `x=2 ` is `11`
Question: (4) Find the slope of the tangent to the curve `y=x^2-3x+2` at the point whose `x- `coordinate is `3`
Solution:
Here `y = x^3-3x +2`
`dy/dx = d/(dx) (x^3-3x +2)`
` 3x^2 - 3`
`[dy/dx]_ (x=3)`
` = 3xx(3)^2 - 3=27 - 3=24`
Thus slope of tangent at `x=3 is 24`
Question: (5) Finfd the slope of the normal to the curve `x=a cos^2 theta, y= a sin^2 theta` at `theta = pi/ 4`
Solution:
Here `x=a cos^2 theta` and ` y= asin ^3 theta`
Differentiating both sides w.r,t, `theta`, we have
`(dx)/(d theta)`
`=d/(d theta) [a cos^3 theta]` and `(dy)/(d theta) `
`= d/(d theta) [a sin ^2 theta]`
` =3a cos^2 theta . (- sin theta)`
`=3a sin^2 theta. (cos theta)`
`= 3a sin theta cos^2 theta`
Now `(dy)/(dx)= (dy//d theta)/(dx//d theta)`
` =(3a sin^2 theta cos theta)/(-3a sin theta cos ^2 theta)`
` =-tan theta`
Slope of normal at `theta = pi/4 is (-1)/ [(dy)/(dx)]_ (theta = pi/4)`
` = -1/[-tan pi/4]`
`= -1/-1 =1`
Question: (6) Find the slope of the normal to the curve `x= 1-a sin theta, y= b cos^2 theta` at `theta =pi/2`.
Solution:
Here ` x= 1 -a sin theta ` and `y = b cos^2 theta`
Differentiating both sides w. r. t. `theta` , we have
`(dx)/(d theta)- d/(d theta)[ 1- a sin theta]` and ` (dy)/(d theta)`
`= d/(d theta) [b cos ^2 theta]`
`=- a cos theta `
` =2b cos theta . (-sin theta)`
`=- 2b sin theta cos theta `
Now `(dy)/(dx)= (dy//d theta)/(dx// d theta)=`
` (-2b sin theta cos theta)/(-a cos theta)`
`( 2b sin theta)/a`
Slope of normal at ` theta = pi/ 2 `is` (-1)/[dy/dx]_ (theta = pi/2)`
`= (-1)/[2b/a sin theta]_( theta =pi/2)`
`= (-a)/(2bsin pi/2) = (-a)/(2b)`
Thus slope of normal at `theta = pi/2 `is ` - a/(2b)`
Question: (7) Find points at which the tangent to the curve `y= x^3 - 3x^2- 3x+ 7` Is parallel to `x-` axis
Solution:
Here ` y = x^2- 3x^2-9x +7`
`(dy)/(dx) = d/(dx)[x^3 - 3x^2 -9x +7]`
` 3x^2 - 6x- 9 `
`=3(x^2-2x-3)`
` 3(x^2- 3x +x -3)`
` 3(x-3) (x+1)`
Slope of tangent `= 3 (x-3) (x+1)`
Since tangent is parallel to `x-` axis,
`:. (x-3)(x+1)=0`
`=> x=3 ` or ` x=-1`
`x=3`
Then `y=(3)^3-3xx(3)^2-9xx3+7`
` 27 - 27 - 17 +7=-20`
` When `x=-1`
Then `y=(-1)^3-3xx(-1)^2-9xx-1+7`
`=-1-3+9+7=12`
Thus required points are `(3, -20)` and `(-1, 12)`
Question: (8) Find a point on the curve ` y=(x-2)^2` at which the tangent is parallel to the chord joining the points `(2, 0)` and `(4,4)`
Solution:
Here ` y= (x-2)^2`
`(dy)/(dx) = d/(dx)[(x-2)^2]`
`= 2(x-2)`
Slope of tangent `= 2(x-2)`
Slope of the chord joining the points `(2, 0)` and `(4, 4)` is
` (4-0)/(4-2)= 4/2 =2`
It is given that tangent is parallel to the chord joining points `(2, 0)` and `(4,4)`=> `slope of tangent = slope of chord`:. 2(x-2)= 2`
` => x-2 =1 => x=3`
When `x=3`
Then `y= (3-2)^2=1`
Thus required point is `(3,1)`
Question: (9) Find the point on the curve `y=x^3 - 11x + 5 ` at which the tangent is `y=x-11`
Solution:
Here ` y= x^3-11x+5`
`(dy)/(dx) = d/(dx)(x^3-11x+5)`
`=3x^2 -11`
Slope of tangent `= 3x^ -11`
Slope of the line `y= x= 11 is 1`
`:. 3x^2 -11 =1 `
`=>3x^2 = 12`
` => x^2 =4`
`=> x=+- 2`
When `x=2`
Then `y 2-11 =-9`
When `x=-2`
Then y=-2-11=-13`
Now point `(-2,-13)` doesnot lie on the curve
Thus the required point on the curve is `(2, -9)`
Question: (10) Find the eqution of all lines having slope`-1` that are tangents to the curve `y= 1/ (x-1)^+ x!= `.
Solution:
Here `y= 1/(x-1)`
` (dy)/(dx) = d/(dx)1/(x-1)`
`=- 1/(x-1)^2`
It is given that slope of tangent is `-1`
`:. - 1/(x-1)^2 `
`-1=> (x-1)^2`
`=1=>x -1 =+-1=>`
` =>x=2` or `x=0`
When `x=2`
Then `y= 1/(2-1)=1`
When `x= 0`
Then `y=1/(0-1)=-1`
The points on the given curve are (2,1) ` and `(0,-1)`
Equation of tangent at points(2, 1)`is
` y-1=-1(x-2)`
`=> y-1 =-x+2`
` => x+y-3=0`
Equation of tangent at point `(0, -1) `is
`y+1 =-(-x-0)`
`=> y+1 =-x`
`=> x+y+1=0`
Thus required lines are `x+y-3 =0` and `x+y+1=0`
Question: (11) Find the equation of all lines having slope 2 which are tangents to the curve ` y= 1/(x-3') x!=3.`
Solution:
Here `y=1/(x-3)`
` (dy)/(dx)= d/(dx)(1/(x-3))`
` =- 1/(x-3)^2`
It is given that slope of tangent is `2`
`:. -1/ (x-3)^2 = 2 `
`2 => (x-3))^2 =- 1/2 `
Which is not possible because square of a number cannot be equal to a negative number.Thus these exist no tangent having slope `2`.
Question: (12) Find the eqution of all lines having slope `0` which are tangent to the curve
` y= 1/(x^2-2x+3)`
Solution:
Here ` y= 1/(x^2-2x+3)`
`(dy)/(dx)= d/(dx) 1/(x^2 - 2x +3)`
`d/(dx)[(x^2-2x +3)^-1]`
` - 1/(x^2-2x +3)^2 xx(2x-2)`
`=(-2 (x-1))/(x^2-2x+3)^2)`
It is given that slope of tangent is `0`
`:. (-2(x-1))/(x^2 -2x + 3)^2`
`=0 => -2(x-1)=0 => x=1`
When `x=1`
Then `y = 1/ ((1)^2-2xx1+3)`
`1/(1-2+3)=1/2`
Thus point on curve is `(1, 1/2)`
Equation of tangent at point `(1, 1/2)` is
` y=1/2 =`
`=0(x-1)=>y-1/2=0`
` =>y=1/2`
Thus required equation of line is `y=1/2`
Question: (13) Find the points on the curve `(x^2)/9 + (y^2)/16 =1` at which tangents are (i) parallel to `x-` axis (ii) parallel to `y-` axis.
Solution:
Here equation of curve is` (x^2)/9+(y^2)/16 =1`
Differentiating both sides w.r.t. `x` we have
`d/(dx)((x^2)/9 + (y^2)/16)`
` = d/(dx)(1)`
`:. (2x)/9 + (2y)/16 (dy)/dx =0`
`=> (dy)/dx`
` = (-2x)/9 xx 16/2y `
` = - (16x)/9y`
(i) tangent is parallel to `x-`axis
`:. (dy)/(dx)=0`
`:. - (16 x)/9y= 0 => x=0`
When `x=0 `then ` (0)^2/9+ y^2/16 =1`
` => y^2=16 => y=+- 4`
Thus required points are `(0,4)`and `(0, -4)`
(ii)tangent is parallel to `y-` axis
`:. (dy)/(dx) = 1/0`
` :. - (16 x)/9y = 1/0 `
`=> 9y = 0 => y=0 `
When `y= 0` then `x^2/9+(0)^2/16 =1`
`=>x^2 = 9 =>`
`x=+-3`
Thus required points are `(3,0)`
Question: (14) Find the eqution of the tangent and normal to the given curves at the indicated points.
(i) `y= x^4-6x^3+ 13 x^2- 10 x + 5 at (0, 5)`
(ii)` y=x^4-6x^3 + 13 x^2 - 10 x + 5at (1,3)`
(iii) `y= x^3 at (1, 1)`
(iv)` y=x^2 at (0,0)`
(v)`x=cos t, y= sin t` at `t = pi/4`
Solution:
(i) Here ` y= x^4 - 6^3+ 13x^2- 10x +5`
` (dy)/(dx)= d/(dx)(x^4-6x^3 +13 x^2 - 10x + 5)`
` = 4x^3 - 18 x^2 + 26 x -10`
`[(dy)/(dx)]_(x=0) `
`= 4xx (0)^3 - 18 xx (0)^2 + 26 xx0 - 10`
` =- 10`
`:.` Equation of tangent at `(0, 5)` is
` y-5 =- 10 (x-0) `
`=> y-5 =-10x `
`10x + y -5 =0`
Equation of normal at `(0,5)` is
` y=5 = 1/10 (x-0) `
`=> 10 y - 50 =x`
`=>x- 10 y + 50=0`
(ii) Here `y=x^4-6x^3 + 13 x^2 - 10 x + 5`
`=4x^3 - 18 x^2 + 26 x -10`
`[(dy)/(dx)]_(x=1) `
`= 4xx (1)^3 - 18 xx (1)^2 + 26 xx1 - 10`
` =4- 18 + 26 - 10 =2`
` :. Equation of tangent at (1, 3)` is
` y-3 = 2(x-1)=> y-3`
` =2x -2 => 2x -y +1 =0`
Equation of normal at `(1, 3) ` is
` y-3 = - 1/2 (x-1)`
`=> 2y - 6 =-x+1`
`=> x+2y-7=0`
(iii) Here `y= x^3`
`(dy)/(dx)= d/(dx) (x^3)`
`=3x^2`
`[(dy)/dx]_(x=1)`
` = 3xx (1)^2 =3`
Equation of tangent at `(1, 1)` is
` y-1 = 3(x-1)`
`=>y-1 = 3x -3`
` => 3x - y -2 =0`
Equation o normal at `(1, 1) is
` y-1 = - 1/3 (x-1)`
`=> 3y -3=-x+1`
` => x+3y-4=0`
(iv) Here ` y=x^2`
` (dy)/(dx) = d/(dx) (x^2)`
`= 2x`
`[(dy)/(dx)]_(x=0)`
` = 2xx (0)=0`
Equation of tangent at `(0, 0)`is
` y-0 = 0 (x - 0)`
`=> y=0`
Equation of normal at `(0, 0)` is
`y=0 = 1/0 (x-0)=> x=0`
(v) Here `x= cos t` and`y=sin t`
Differentiating both sides w.r.t. `t` we have
`(dx)/(dt)= d/(cos t)` and ` (dy)/(dx) = d/(dt) (sin t)`
`=- sin t = cos t`
Now `(dy)/(dx)`
`= ((dy)//(dt))/((dx)//(dt))`
` (cos t)/(-sin t)`
` =- cot t`
`[(dy)/(dx)]_(t=pi/4)`
` =-cot (pi/4) -1`
At `t=pi/4`
`x= cos pi/4`
` = 1/sqrt 2 ` and
`y = sin pi/4 = 1/sqrt 2`
`:.` Equation of tangent at `( 1/sqrt2 , 1/ sqrt 2)` is
`y- 1/sqrt2 =- 1(x-1/sqrt 2)`
` =>y-1/sqrt 2`
` =- x +1/sqrt2`
` => x+y -y- sqrt2=0`
Equation of normal at `( 1/sqrt2. 1/sqrt2)` is
`y -1/sqrt 2 `
`y- 1/sqrt2`
` = -1/ -1(x- 1/sqrt 2)`
`=> y- 1/sqrt 2`
` =x- 1/ sqrt2`
`=> x=y`
Reference: