Chapter 6: Application of Derivatives

Given, a spherical ball of salt

Then, the volume of ball V = 4/3 πr³ where r = radius of the ball

Now, according to the question we have

dV/dt ∝ S, where S = surface area of the ball

Therefore, the radius of the ball is decreasing at constant rate.

We know that the area of circle, A = πr², where r = radius of the circle

And, perimeter = 2πr

According to the question, we have

Therefore, it’s seen that the perimeter of the circle varies inversely as the radius of the circle.

Given,

Height of the kite(h) = 151.5 m

Speed of the kite(V) = 10 m/s

Let FD be the height of the kite and AB be the height of the kite and AB be the height of the boy.

Now, let AF = x m

So, BG = AF = x

And, dx/dt = 10 m/s

From the figure, it’s seen that

GD = DF – GF = DF – AB

= (151.5 – 1.5) m = 150 m [As AB = GF]

Now, in ∆ BDG

BG² + GD² = BD² (By Pythagoras Theorem)

x² + (150)² = (250)²

x² + 22500 = 62500

x² = 62500 – 22500 = 40000

x = 200 m

Let initially the length of the string be y m

So, in ∆ BDG

BG² + GD² = BD²

x² + (150)² = y²

Differentiating both sides w.r.t., t, we have

Therefore, the rate of change of the length of the string is 8 m/s.

Let’s consider P to be any point at which the two roads are inclined at an angle of 45^o.

Now, two men A and B are moving along the roads PA and PB respectively with same speed ‘V’.

And, let A and B be their final positions such that AB = y

∠APB = 45^o and they move with the same speed.

So, ∆APB is an isosceles triangle.

Now, draw PQ ⊥ AB.

We have, AB = y

So, AQ = y/2 and PA = PA = x (assumption)

And, ∠APQ = ∠BPQ = 45^o/2 = 22.5^o

Now, in right ∆APQ

sin 22.5^o = AQ/AP

Therefore, the rate of their separation is

According to the question, we have

Therefore, the required angle is π/3.

(1.999)⁵ = (2 – 0.001)⁵

Let x = 2 and ∆x = -0.001

Also, let y = x⁵

Differentiating both sides w.r.t, x, we get

dy/dx = 5x⁴ = 5(2)⁴ = 80

Now, ∆y = (dy/dx). ∆x = 80. (-0.001) = -0.080

And, (1.999)⁵ = y + ∆y

= x⁵ – 0.080 = (2)⁵ – 0.080 = 32 – 0.080 = 31.92

Therefore, approximate value of (1.999)⁵ is 31.92

Given,

The internal radius r = 3 cm

And, external radius R = r + ∆r =3.0005 cm

∆r = 3.0005 – 3 = 0.0005 cm

Let y = r³ ⇒ y + ∆y = (r + ∆r)³ = R³ = (3.0005)³

Differentiating both sides w.r.t., r, we get

Therefore, the approximate volume of the metal in the shell is 0.018π cm³

Let AB is the height of street light post and CD is the height of the man such that

AB = 5(1/3) = 16/3 m and CD = 2 m

Let BC = x length (the distance of the man from the lamp post)

And CE = y is the length of the shadow of the man at any instant.

It’s seen from the figure that,

∆ ABE ~ ∆ DCE [by AAA similarity criterion]

Now, taking ratio of their corresponding sides, we have

Differentiating both sides w.r.t, t, we have

So, the length of shadow is decreasing at the rate of 1 m/s.

Now, let u = x + y

(where, u = distance of the tip of shadow from the light post)

On differentiating both sides w.r.t. t, we get

Therefore, the tip of the shadow is moving at the rate of m/s towards the light post and the length of shadow decreasing at the rate of 1 m/s.

Given, L = 200(10 – t)² where L represents the number of liters of water in the pool.

On differentiating both the sides w.r.t, t, we get

dL/dt = 200 x 2(10 – t) (-1) = -400(10 – t)

But, the rate at which the water is running out

= -dL/dt = 400(10 – t)

Now, rate at which the water is running after 5 seconds will be

= 400 x (10 – 5) = 2000 L/s (final rate)

T = 0 for initial rate

= 400 (10 – 0) = 4000 L/s

So, the average rate at which the water is running out is given by

= (Initial rate + Final rate)/ 2 = (4000 + 2000)/ 2 = 6000/2 = 3000 L/s

Therefore, the required rate = 3000 L/s

Let’s assume x to be the length of the cube.

So, the volume of the cube V = x³ …. (1)

Given that, dV/dt = K

Now, on differentiating the equation (1) w.r.t. t, we get

dV/dt = 3x². dx/dt = K (constant)

So, dx/dt = K/3x²

Now,

Surface area of the cube, S = 6x²

Differentiating both sides w.r.t. t, we get

Therefore, the surface area of the cube varies inversely as the length of the side.