Application of Derivatives NCERT Solutions

NCERT Exercise 6.5:Q:24-29

Question: (24)Show that the right circular cone of least curved surface and given volume has an altitude equal to `sqrt 2 ` times the radius of the base.

Solution:

Let `r` and `h` be the radius of base and altitude respectively of a cone then

`V = 1/3 pir^2h = k`

`=> h=3K/(pi r^2)`

Let `S` denote the surface areaof cone then.

` S = pirl = pir sqrt(h^2+r^2)`

From (i)and (ii),we have

`S = pi r sqrt((9k^2)/(pi^2r^4)) +r^2 )`

`=(pir)/(pi r^2)sqrt(9h^2+pi^2r^6)`

`1/r sqrt (9k^2+pi^2 r^6)`

For maxima or minima.`(dS)/(dr)=0`

`:. (2pi^2r^6-9k^2)/(r^2sqrt(9k^2+pi^2r^6))=0`

`=> 2pi^2r^6-9k^2=0`

`9k^2 = 2pi^2 r^6`

`=> k^2 = 2/9 pi^2r^6`

and `r^6=(9k^2/2pi^2)`

`r= [(9k^2)/(2pi^2)]^(1/6)`

At `r= [(9k^2)/(2pi^2)]^(1/6)`

When `r < r= [(9k^2)/(2pi^2)]^(1/6)`

then `(dS)/(dr)` is `- Ve`

When `r > [(9k^2)/(2pi^2)]^(1/6)`

then `(dS)/(dr)` is `+ Ve`

`r =[(9k^2)/(2pi^2)]^(1/6)` is a point of minima

now from (i) ` k=(pir^2h)/3`

` k^2 = (pi^2r^4h^2)/9 `

`:. (2pi^2 r^6)/9`

`=(pi^2r^4h^2)/9`

`[because; k^2=(2pi^2r^6)/9]`

`:. h^2=2pi^2 => h= sqrt 2r`.

Question: (25)Show that the semi - vertical angle of the cone of the maximum volume and of given slant height is `tan^-1 sqrt2` .

Solution:

Let theta be the semi- vertical angle of the cone of height `h`and slant height `l`

Then `CD= l cos theta` and `BD=lsin theta`

Let `V` denote the volume of cone then

`V = 1/3 pi . BD^2.CD`

` =1/3 pi (lsin theta)^2. (l cos theta)`

`= pi/3 (l^3sin ^2 theta cos theta)`

`:. (dV)/(d theta)= (pi l^3 )/3 [sin^2 theta (-sin theta) + ` `cos theta(2sin theta cos theta)]`

`=(pil^3)/3 [-sin^3 theta + 2 sin theta cos ^2 theta]`

`= (pi l^3)/3 sin theta [2 cos^2 theta-sin^2 theta ]`

`= (pi l^3)/3 cos ^3 theta [ 2-7 tan^2theta]`

For maxima or minima, `(dV)/(d theta)=theta`

`:. (pi l^3)/3 sin theta (2 cos^2 theta - sin^2 theta)=0`

`=> sin theta (2 cos^2 theta - sin^2 theta) =0`

` => 2 cos^2 theta - sin^2 theta =0`

` => sin theta ( 2 cos^2 theta - sin^2 theta)=0`

`=> 2 cos ^2 theta - sin ^2 theta =0`

`=> 2 cos^2 theta = sin ^2 theta `

`=> tan^2 theta = 2 => tan theta = sqrt 2`

Now ` cos theta = 1/(sec theta)`

`= 1/ (sqrt(1+tan^2 theta))`

`=1/(sqrt(1+2)`

`=1/(sqrt3)`

At ` theta = tan ^-1 sqrt 2`

`(d^V)/(d theta^2)`

`=(pi l ^3)/3 (1/sqrt 3)^2 (2- 7 xx 2)`

` (pi l^3)/(9sqrt3) xx -12`

` = - 4/(3 sqrt 3) pi l ^3 < 0`

Thus volume is maximum when semi- vertical angle of cone is `tan^-1 sqrt 2`

Question: (26)-Show that semi- vertical angle of right circular cone of given surface area and maximum volume is `sin^-1 (1/3)`

Let `r` be the radious of base, `h` be the height and `l` be the slant height of cone.

Then `S= pirl + pir^2 =K`

`=>l = (K-pir^2)/(pir)`----(i)

and `l= sqrt(h^2 + r^2)`

Let `V` denote the volume of cone then

`V=1/3 pir^2 h`

`V =1/3 pi r^2 sqrt (l^2 -r^2) `

`= 1/3 pir^2 sqrt((K-pi r^2)/(pir)-r^2`

`:. v^2 = (pi^2 r^4)/9[((K-pir^2)^2-pi^2r^4)/(pi^2r^2)]`

`(pi^2 r^4)/9[(K^2 + pi^2r^4-2piKr^2-pi^2r^4)/(pi^2r^2)]`

`(pi^2 r^4)/9[(K^2-2pi Kr^2)/(pi^2r^2)]`

`= (pi^2 r^4)/9 xx K/(pi^2r^2)(K-2pir^2)`

` V^2 = (Kr^2)/9(K - 2pir^2)`

` K/9 [Kr^2- 2pir^4]`

`(d(V^2))/(dr)`

` K/9[2Kr-8pir^3]`

`(d^2(v^2))/(dr^2)`

`K/9[2K-24pir^2]`

For maxima or minima, `(dV^2)/(dr)=0`

`:. K/9 [2Kr-8pir^3]`

`=> 2Kr-8pir^3=0`

`=> 8pir^3 = 2Kr`

`=> r^2 = K/(4pi)`

`=>r = sqrt(K/(4pi))`

At `r= sqrt(K/(4pi))`

`(d^2(V))/(dr^2) `

`=K/9 [2K-24pi xx K/(4pi)]`

` =K/9[2K- 6K]`

` =-(4k^2)/9<0`

`:. r = sqrt K/(4pi) ` is a point of maxima.

Now ` pirl + pir^2 = K `

`=> pirl + pir^2 = 4pir^2`

`=> `pirl = 3 pir^2`

Also `sin theta = r/l = r/(3r) = 1/3`

`=> theta = sin ^-1 (1/3)`

Thus volume is maximum when semi- vertical angle is ` sin^-1 (1/3)`

Choose the correct answr in the Exercises `27` to `29`.

Question: (27) The point on the curve `x^2 = 2y` which is nearest to the point `(0,5)` is

Question: (A) `(2sqrt2.4)`

Question: (B)`(2sqrt 2 .0)`

Question: (C)`(0,0)`

Question: (D) `(2,2)`

Let `A (x,y) ` be any point on the given curve and `B(0,5) ` be given point then

`S = AB^2 = (x-0)^2+ (y-5)^2`

` = x^2+ y^2 + 25 - 10y`

`=2y+y^2+25-10y`

` =y^2-8y+ 25`

`:. (ds)/(dy)=2y-8`

`(d^2s)/(dy^2) =2`

For maxima or minima , `(ds)/(dy)=0`

`:. 2y-8=0 `

`=>y = 4`

At `y= 4`

`(d^2s)/(dy^2) =2 > 0 `

` y=4` is a point of minima.

now `x^2 = 2y => x^2`

`2 xx 4 => x`

` sqrt 8=2sqrt2`

Thus required point is `( 2sqrt2, 4)`

So answer is (A).

Question: (28)For all real values of x, the minimum value of `(1-x+x^2)/(1+x+x^2)` is

Question: (A) 0

Question: (B) 1

Question: (C) 3

Question: (D) `1/3`

Let `y= (1-x+x^2)/(1+x+x^2)`

`=(-8x^3+12x+4)/(1+x+x^2)^3 `

`= -4(2x^3 - 3x -1)/(1+x+x^2)^3`

for maxima or minima, `(dy)/(dx)=0`

`:. (2(x+1)(x-1)/(1+x+x^2)^2 = 0`

`=> 2(x+1)(x-1)=0`

`=>x=1 or x=1`

At ` x=-1`

`(d^2y)/(dx^2)`

`(-4[2(-1)^3-3 xx -1 -1])/([1-1+(-1)^2]^3)`

` (-4[-2+3-1])/1 = 0`

`:. x=-1` is a point of inflecxion

At `x=1`

`(d^2y)/(dx^2) `

`=(-4[2(1)^3- xx -1-1])/[1+1+(1)^2]^3`

` =(-4[2-3-1])/27`

`= (-4 xx-2)/27 = 8/27`

`:.x =1 ` is a point of minima

`:.` Minimum value `=(1-1+(1)^2)/(1+1+(1)^2)= 1/3`

Thus answer is (D),

Question: (29)The maximum value of ` [x(x-1)+1]^(1/3) , 0 <- x <- 1` is

Question: (A)`(1/3)^(1/3)`

Question: (B)` 1/2`

Question: (C)1

Question: (D)0

Let ` y= [x(x-1) + 1 ]1/3`

`(dy)/(dx) = 1/3 [x(x-1)+ 1] ^-(2/3).(2x-1)`

`= (2x-1)/(3[x (x-1)+ 1]^(2/3))`

`(-2x^2+ 2x +4)/(9[x(x-1)+1)^(5/3))`

`(-2(x^2-x-2))/(9[x(x-1)+1)^(5/3))`

For maxima or minima, `(dy)/(dx)=0`

`:.(2x-1)/(3[x(x-1)+1])^(2/3)= 0`

`=>2x-1 = 0`

`=> x = 1/2`

At `x=1/2`

`(d^y)/(dx)^2`

` = (-2[(1/2)^2- 1/2 -2])/(9[1/2(1/2-1)+1]^(5/3))`

`= (-2[1/4 -1/2 -2])/(9[1/4+1]^(5/3)`

`(-2[- 9/4])/(9(5/4))^(5/3)`

`= (9/2)/(9(5/4)^(5/3))>0`

`:. x= 1/2` is a point of minima.

At `x=0`

`y= [0(0-1)+1](1/3)`

`=(1)^(1/3) =1`

At `x=1

`y= [1(1-1)+ 1]^(1/3)`

`= (1)^(1/3)=1`

Thus maximum value of `y` is 1

So correct answer s (C).

12-math-home

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