Application of Derivatives

Let x be the depth of the wheat inside the cylindrical tank is with radius = 10 cm

Then, Volume V of the cylindrical tank is

V = π (10)²x = 100πpx m³

As $\frac{d}{dt}=314\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$

$\Rightarrow 100\pi \frac{dx}{dt}=314$

$\frac{dx}{dt}=\frac{314}{100\pi }=\frac{314}{100\pi 3.14}=\frac{314}{314}=1.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ i e, rate of increasing of depth

option (A) is correct

Kindly go through the solution

Kindly go through the solution

We have,

f (x) defined on [a, b]

And f (x) > 0 ∀ x ∈ [a, b].

Let x₁, x₂ ∈ [a, b] and x₂>x₁

In the internal x₁, x₂], f (x) will also be continuous and differentiable.

Hence by mean value theorem, there exist c [x₁, x₂] such that

$f(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_{1.}}$

f (x) > 0 ∀ x ∈ [a, b].

Then, f (c) > 0.

$\Rightarrow \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}>0$

i.e., f (x₂) -f (x₁) > 0

f (x₂) >f (x₁).

Hence, the function f (x) is always increasing on [a, b]

Let x be the radius of the sphere ad x be radius of the right circular cone.

Let height of cone = y

Then, in ΔOBA,

(y-r)² + x² = r²

y² + r²- 2ry + x² = r²

x²=2ry - y²

So, the volume V of the cone is

$V=\frac{1}{3}\pi ·x^2·y$ $\left\{=\frac{1}{3}\pi {(radius)}^2*hight\right\}$

$=\frac{1}{3}*\left(2xy-y^2\right)y$

$=\frac{1}{3}x\left(2x{y}^2-y^3\right).$

So, $\frac{dv}{dy}=\frac{\pi }{3}\left[4xy-3y^2\right]$

And $\frac{d^2}{d{y}^2}=\frac{x}{3}[4x-6y].$

At $\frac{dV}{dy}=0.$

$\Rightarrow \frac{\pi }{3}\left[4xy-3y^2\right]=0$

4x -y- 3y² = 0

$\Rightarrow y=\frac{4x}{3}$ asy> 0.

At y = $\frac{4\pi }{3},\frac{d^{2}v}{d{y}^2}=\frac{\pi }{3}[4x-8x]$ = $-\frac{4\pi r}{3}<0.$

Ø V is maximum when y = $\frac{4\pi }{3}$

We have,

f (x) = cos²x + sin x, x∈ [0, π ].

So, f (x) = 2 cos x ( -sin x) + cos x = cos x (1 - 2 sin x).

At f (x) = 0

cosx (1 - 2 sin x) = 0

cosx = 0  or  1 - 2 sin x = 0

cosx = cos $\frac{\pi }{2}$ or sin x = $\frac{1}{2}$ = sin $\frac{\pi }{6}$ = sin $x-\frac{\pi }{6}$

x= $\frac{\pi }{2}$ , x = $\frac{\pi }{6}$ and x = $\frac{5\pi }{6}$  [0, π ].

So, f $\left(\frac{\pi }{2}\right)$ = cos²$\frac{\pi }{2}$ + sin $\frac{\pi }{2}$ = 1.

Absolute minimum of f (x) = $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and absolute minimum of f (x) = 1.

Let P be the point on hypotenuse of a triangle. ABC, t angle at B.

Which is at distance a& b from the sides of the triangle.

Let < BAC = < MPC = .

Then, in… ΔANP,

$\Rightarrow ta{n}^3\theta =\frac{a}{b}$

$tan\theta ={\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..}$

At tan Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3,$}\right.}$

$\frac{d^2\stackrel{?}{q}}{d{\theta }^2}=+ve.>0.$ { Øall trigonometric fxⁿ are + ve in I^st quadrant}.

So, z is least for tanØ = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$

As, Sec²Ø = 1 + tan²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$ = $\frac{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}}$

secØ = ${\left[\frac{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = $\frac{{\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..}}$

And tan²Ø = $\frac{1}{co{t}^2\theta }$

cot²Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

And cosec²Ø = 1 + cot²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ = $\frac{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

cosecØ = $\frac{{\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..}}{a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

Let 'x' metre be the radius of the semi-circular opening mounded on the length '2x' side of rectangle. Then, let 'y' be the breadth of the rectangle.

Then, perimeter of the window = 10m

$\Rightarrow \frac{2\pi x}{2}+2x+y+y=10$

x + 2x + 2y + = 10.

$\Rightarrow y=\frac{10-(\pi +2)\pi }{2.}$

Let the area of the window be A.

Then, A = $\frac{1}{2}\left(\pi x^2\right)+2xy.$

$=\frac{\pi x^2}{2}+2x·\left[\frac{10-(x+2)x}{2}\right].$

$=\frac{\pi x^2+20x-2\pi x^2-4x^2}{2}$

= $\frac{1}{2}$ [-πx²- 4x² + 20x].

So, $\frac{d}{dx}=\frac{1}{2}$ [ -2πx - 8x + 20]

And $\frac{d\stackrel{2}{}}{d{x}^2}=\frac{1}{2}$ [ -2π - 8] = -π -4 = -( π+ 4)

At $\frac{d}{dx}=0.$

$\frac{1}{2}$ [ -2πx - 8x + 20] = 0

2x + 8x = 20

x = $\frac{20}{2\pi +8}$ = $\frac{10}{\pi +4}.$

At x = $\frac{10}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}$ = -( π+ 4) < 0

Øx = $\frac{10}{x+4}$ is a point of minima.

And y = $\frac{10-(x+2)*\left(\frac{10}{x+4}\right)}{2}$

$=\frac{10(\pi +4)-(\pi +2)10}{2(\pi +4)}$

$=\frac{10\pi +40-10\pi -20}{2(x+4)}=\frac{10}{\pi +4.}$

Ø Dimensions of the window are

length = 2x = $\frac{20}{\pi +4}m$

breadth = y $\frac{10}{\pi +4}m.$

radius = y = $\frac{10}{\pi +4}m.$

Let r and s be the radius of the circle and length of side of square.

Then, sum of perimeter of circle and square = k

2πr +4s = k

s = $\frac{k-2*r}{4}$

The area A be the total areas of the circle and square.

Then, A = πr² + s²

$=\pi r^2+{\left(\frac{k-2\pi r}{4}\right)}^2=\pi r^2+\frac{x^2+4{\pi }^2{r}^2-4\pi rk}{16}$

$=\frac{16\pi r^2+x^2+4{\pi }^2{r}^2-4\pi k\cdot r}{16}$

$=\frac{1}{16}\left[\left(16\pi +4{\pi }^2\right){\pi }^2-4\pi kr+x^2\right].$

So, $\frac{dA}{dr}=\frac{1}{16}\left[\left(16\pi +4x^2\right)\cdot 2n-4\pi k.\right].$

And $\frac{d^{\mathrm{2A}}}{d{r}^2}=\frac{1}{16}\left[\left(6\pi +4{\pi }^2\right)2\right]=\frac{16\pi +4{\pi }^2}{8}$

At $\frac{dA}{dr}=0$

$\frac{1}{16}\left[\left(16x+4x^2\right)2x-4\pi k\right]=0$

$\Rightarrow \left(16x+4x^2\right)2x-4\pi k=0$

$\Rightarrow 4\pi [4+\pi ]2r=4\pi k.$

$r=\frac{k}{2(4+\pi ).}$

At $r=\frac{x}{2(4+\pi )},\frac{d^{\mathrm{2A}}}{d{x}^2}=\frac{16\pi +4{\pi }^2}{8}>0.$

s = $\frac{2x}{2(4+x).}$

s = 2r.

Hence, proved.