Application of Derivative Exercise: 18.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 18.2

1. a) A particle moves in a straight line. The distance s covered by the particle in time t is given by s = 2t² + 5t – 4.

Solution:

Given, s = 2t² + 5t – 4

Or, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 4t + 5 and $\frac{{{{\rm{d}}^2}{\rm{s}}}}{{{\rm{d}}{{\rm{t}}^2}}}$ = 4.

At time(t) = 6 seconds.

Velocity (v) = $\frac{{{\rm{dS}}}}{{{\rm{dt}}}}$ = (4 * 6 + 5) = 29 m/sec.

Acceleration = $\frac{{{{\rm{d}}^2}{\rm{S}}}}{{{\rm{d}}{{\rm{t}}^2}}}$ = 4m/sec². 

b) The displacement of the particle varies with time according to the relation x = – 15t² + 20t + 3. Find the velocity and the acccleration of the particle in ½ second. The distance is measured in metere and time in second.

Solution:

Given, x = – 15t² + 20t + 30

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = – 30t + 20; $\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}}$ = –30.

When time (t) = $\frac{1}{2}$ second, velocity $\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)$ = –30 * $\frac{1}{2}$ + 20 = 5 m/sec.

Acceleration $\left( {\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{t}}^2}}}} \right)$ = – 30 m/sec².

 

2. a) The side of a square sheet is increasing at the rate of 5cm/min. At what rate is the area increasing when the side is 12cm long?

Solution:

Let x be the side and A be the area of a sheet at time t.

Given. $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = 5cm/min., $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = ?

We have, A = x².

Or, $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = 2x.$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$

When x = 12cm, $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = 2 * 12 * 5 = 120cm²/mic.

So, the area increasing at the rate 120cm²/min.

b) A stone thrown into a pond produces a circular ripples which expands from the point of impact. If the radius of the ripple increases at the rate of 3.5cm/sec, how fast is the area growing when the radius is 15 cm?

Solution:

Let r be the radius and A be the area of the circular ripples at time t.

Given. $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ = 3.5cm/min., $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = ?

We have, A = πr².

Or, $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = 2πr.$\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$

When r = 15cm, $\frac{{{\rm{dA}}}}{{{\rm{dt}}}}$ = 2π * 15 * 3.5 = 105 * $\frac{{22}}{7}$ = 330cm²/sec.

 

3. a) From a cylindrical drum containing oil and kept vertical, the oil is leaking so that the level of the oil is decreasing at the rate of 2cm/min. If the radius and the height of the drum is 10.5 cm and 40cm respectively. Find the rate at which the volume of the oil is decreasing.

Solution:

Let r be the radius, h be the height and v the volume of the cylindrical drum at time t.

Given, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = 2cm/min.

r = 10.5cm, h = 40cm, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = ?

We know, v = πr²h

Or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = πr².$\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$  [So, r is constant]

Where r = 10.5, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = $\frac{{22}}{7}$ * (10.5)² * 2 = 693 cm³/min.

So, the volume decreasing at the rate of 693 cm²/min.

b) Water is poured into a right circular cylinder of radius 8cm at the rate of 18cu.cm/min. Find the rate at which the level of water is rising in the cylinder.

Solution:

Let r be the radius, h be the height and v the volume of the volume of the circular cylinder at time t.

Given,

r = 8cm, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 18cu.cm/min. , $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = ?

We know, v = πr²h

Or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = πr².$\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$  [So, r is constant]

Where, r = 8cm, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$= 18 cu.cm/min.

Then, 18 = π * 8² * $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ ⇒ $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = $\frac{{18}}{{64{\rm{\pi }}}}$ = $\frac{9}{{32{\rm{\pi }}}}$

So, the level of water rising at the rate of $\frac{9}{{32{\rm{\pi }}}}$ cm/min.

 

c) Gasoline is pumped into a vertical cylindrical tank at the rate of 24cu.cm/min. The radius of the tank is 9cm. How fast is the surface rising?

Solution:

Let r be the radius, h be the height and v the volume of the cylindrical tank at time t.

Given,

$\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 24 cu.cm/min., r = 9cm , $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = ?

We know, v = πr²h

Or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = πr².$\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$  [So, r is constant]

Or, 24 = π * 9² * $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$.

So, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = $\frac{8}{{27{\rm{\pi }}}}{\rm{\:}}$ cm./min.

 

4. a) A spherical balloon is inflated at the rate of 18cu.cm/sec. at what rate is the radius is increasing when the radius is 8cm?

Solution:

Let r be the radius, v be the volume of a spherical balloon at time t.

Given, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 18 cu.cm/sec, r = 8cm, $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ = ?

We have, v = $\frac{4}{3}$πr³.

Or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = $\frac{{4{\rm{\pi }}}}{3}$.3r².$\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$.

When $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 18 cu.cm/sec and r = 8cm.

Or, 18 = 4π * 8² * $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$.

Or, $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ = $\frac{9}{{128{\rm{\pi }}}}$ cm/sec.

b) A spherical ball of salt is dissolving in water in such a way that the rate of decrease in volume at any instant is proportional to the surface. Prove that the radius is decreasing at the constant rate.

Solution:

Let s be the surface area, r be the radius, v be the volume of the spherical balloon of salt in time t.

We have, v = $\frac{4}{3}$πr³ and s = 4πr².

Or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 4πr².$\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$.

By question, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = –ks (where k is constant).

Then, – ks = s.$\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ ⇒ $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ = –k.

So, the radius is decreasing at a constant rate.

 

5. a) The radius of the conical tank is 1/3 of the height. Water flows into an inverted conical tank at the rate of 4.4cu.cm/sec. How fast the level is rising when the height of the water is 3.5cm?

Solution:

Let r be the radius, h be the height and v be the volume of the conical tank at time t.

Given, r = $\frac{{\rm{h}}}{3}$, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 4.4 cu.cm./sec.

h = 3.5cm, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = ?

We have, v = $\frac{1}{3}$πr²h.

v = $\frac{1}{3}$ * π * ${\left( {\frac{{\rm{h}}}{4}} \right)^2}$* h = $\frac{{{\rm{\pi }}{{\rm{h}}^3}}}{{27}}$

or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = $\frac{{3{\rm{\pi }}}}{{27}}$h².$\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$

When = 3.5cm.

Or, 4.4 = $\frac{{22}}{{7{\rm{*}}9}}$ * (3.5)² * $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$.

Or, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = $\frac{{4.4{\rm{*}}7{\rm{*}}9}}{{22{\rm{*}}12.25}}$ = $\frac{{277.2}}{{269.5}}$ = $\frac{{2772}}{{2695}}$ = $\frac{{36}}{{35}}$.

So, the level rising at the rate of $\frac{{36}}{{35}}$ cm/sec.

b) Water flows into an inverted conical tank at the rate of 42cm³/sec. When the depth of water is 8cm, how fast the level rising? Assume that the height of the tank is 12cm and the radius of the top is 6cm.

Solution:

Let ABC be the conical water tank into which water is flowing at the rate of 42 cm³/sec. at time t. Let h be the height AE of the water and r be the radius EF of the water surface.

Since, $\Delta $ACD ~$\Delta $AFE. So, $\frac{{{\rm{AE}}}}{{{\rm{AD}}}} = \frac{{{\rm{EF}}}}{{{\rm{DC}}}}$.

Or, $\frac{{\rm{h}}}{{12}}$ = $\frac{{\rm{r}}}{6}$ ⇒ r = $\frac{{\rm{h}}}{2}$.

Let v be the volume of the water, then,

v = $\frac{1}{3}$ πr²h = $\frac{1}{3}$π$\frac{{{{\rm{h}}^2}}}{4}$.h = $\frac{{{\rm{\pi }}{{\rm{h}}^3}}}{{12}}$.

or, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = $\frac{{3{\rm{\pi }}{{\rm{h}}^2}}}{{12}}$.$\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$

or, 42 = $\frac{{{\rm{\pi }}{{\rm{h}}^2}}}{4}.\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$.

When h = 8 cm

Or, 42 = $\frac{{{\rm{\pi *}}{8^2}}}{4}{\rm{*}}\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$

Or, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = $\frac{{42{\rm{*}}4}}{{64{\rm{\pi }}}}$ = $\frac{{21}}{{81{\rm{\pi }}}}$

So, $\frac{{{\rm{dh}}}}{{{\rm{dt}}}}$ = $\frac{{21}}{{8{\rm{\pi }}}}$ cm./sec.

 

6) If the volume of the expanding cube is increasing at the rate of 24cm³/min, how fast is its surface area increasing when the surface area is 216 cm²?

Solution:

Let v be the volume, s be the surface area x be the side of the cube at time t.

or,s = 216cm², $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = ?

or, 6x² = 216.

Or, x² = 36.

So, x = 6.

Given, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 24cm³/min.
We have, v = x³.

So, $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$ = 3x².$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$.

Or, 24 = 3x². $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{8}{{{{\rm{x}}^2}}}$

Again, s = 6x².

Or, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 12x.$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = 12x * $\frac{8}{{{{\rm{x}}^2}}}$ = $\frac{{96}}{{\rm{x}}}$ = $\frac{{96}}{6}$

So, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 16cm²/min.

 

7) Two concentric circle are expanding in such a way that the radius of the inner circle is increasing at the radius of the inner circle is increasing at the rate of 10cm/sec and that of the outer circle at the rate of 7cm/sec. At a certain time, the radii of the inner and outer circles ate respectively 24cm and 30cm. At what time, the area between the circles increasing or decreasing. How fast?

Solution:

Let r₁ and r₂ be the radii of two concentric circles s₁ and s₂respectively and let s be the area between two circle at time t.

Given, $\frac{{{\rm{d}}{{\rm{r}}_1}}}{{{\rm{dt}}}}$ = 10cm/sec., $\frac{{{\rm{d}}{{\rm{r}}_2}}}{{{\rm{dt}}}}$ = 7cm/sec.

r₁ = 24cm and r₂ = 30cm.

s = s₂ – s₁ = πr₂² – πr₁²

So, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 2πr₂.$\frac{{{\rm{d}}{{\rm{r}}_2}}}{{{\rm{dt}}}}$ – 2πr₁.$\frac{{{\rm{d}}{{\rm{r}}_2}}}{{{\rm{dt}}}}$.

Where, r₁ = 24cm and r₂ = 30cm

Then, $\frac{{{\rm{dr}}}}{{{\rm{dt}}}}$ = 2π(30 * 7 – 24 * 10) = 2π(–30) = – 60π.

So, they are between two concentric circle decreasing at the rate of 60π cm²/sec.

 

8) A man of height 1.5m walks away from a lamp post of height 4.4m at the rate of 20cm/sec. How fast is the shadow lengthening when the man is 42 cm from the post?

Solution:

Let AB be the lamp, DE on the height of the man, Let E be the position of the man in time t.

Let BE = x, EF = y

Since, $\Delta $ABF ~$\Delta $DEF, then,

Or, $\frac{{{\rm{AB}}}}{{{\rm{DE}}}}$ = $\frac{{{\rm{BF}}}}{{{\rm{EF}}}}$

Or, $\frac{{4.5}}{{1.5}} = \frac{{{\rm{x}} + {\rm{y}}}}{{\rm{y}}}$

Or, x + y = 3y ⇒  x = 2y ⇒ y = $\frac{{\rm{x}}}{2}$.

Given, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = 20cm/sec.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{1}{2}.\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{1}{2}$ * 20 cm/sec. = 10 cm/sec.

 

9) A point is moving along the curve y = 2x³ – 3x² in such a way that its x-coordinate is increasing at the rate of 2cm/sec. Find the rate at which the distance of the point from the origin is increasing when the point is at (2,4).

Solution:

Let P(x,y) be any point along the curve y = 2x³ – 3x² at time t.

Given, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = 2cm/sec. ,(x,y) = (2,4)

Let OP = D.

Then D² = (x – 0)² + (y – 0)² = x² + y².

= x² + (2x³ + 3x²)²

When P is at (2,4), then,

D² = 2² + (2.2³ – 3.2²)² = 4 + (16 – 12)² = 20.

So, D = $\sqrt {20} $ = 2$\sqrt 5 $.

Or, $\frac{{{\rm{d}}{{\left( {\rm{D}} \right)}^2}}}{{{\rm{dt}}}}$ = 2x.$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ + 2(2x³ – 3x²).(6x² – 6x).$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$.

Or, 2D. $\frac{{{\rm{d}}\left( {\rm{D}} \right)}}{{{\rm{dt}}}}$ = [2.2 + 2(2x³ – 3x²)(6x² – 6x)]$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$.

When (x,y) = (2,4),

Or, 2 * 2$\sqrt 5 $ * $\frac{{{\rm{d}}\left( {\rm{D}} \right)}}{{{\rm{dt}}}}$ = [2x + 2(2x³ – 3.2²)(6.2² – 6.2)]2

Or, 4$\sqrt 5 $.$\frac{{{\rm{d}}\left( {\rm{D}} \right)}}{{{\rm{dt}}}}$ = [4 + 2 * 4 * 12]2

Or, $\frac{{{\rm{d}}\left( {\rm{D}} \right)}}{{{\rm{dt}}}}$ = $\frac{{200}}{{4\sqrt 5 }}$

So, $\frac{{{\rm{d}}\left( {\rm{D}} \right)}}{{{\rm{dt}}}}$ = 10$\sqrt 5 $ cm/sec.

 

10. a) A kite is 24m high and there are 25 meters of cord out. If the kite moves horizontally at the rate if 36km/hr directly away from the person who is flying it, how fast is the cord out?

Solution:

Let B be the position of the kite moving horizontally in time t.

Let AB = x, OB = s

OB² = OA² + AB²

s² = 24² + x² …(i)

When s = 25,

Or, 25² = 24² + x²

Or, x² = 25² – 24² = 625 – 576 = 49

So, x = 7.

From equation(i), 2x. $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = 2x.$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$.

When, s = 25, x = 7, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 36 km/hr

Then, 2 * 25. $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = 2 * 7 * 36.

Or, $\frac{{{\rm{ds}}}}{{{\rm{dt}}}}$ = $\frac{{2{\rm{*}}7{\rm{*}}36}}{{2{\rm{*}}25}}$ = 10.08 km/hr.

b) A 2.5m ladder leans against the wall. If the top slides downwards at the rate of 12cm/sec, find the speed of the lower end when it is 2m from the wall.

Solution:

Let AB be the position of the ladder such that:

Let AB = x, OB = y at time t.

Given, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = – 12cm/sec = $ - \frac{{12}}{{100}}$m/sec.

So, x² + y² = (2.5)². ….(i)

When x = 2m.

Or, 2² + y² = 6.25

Or, y² = 2.25

So, y = 1.5m

Now, differentiating (i) w.r.t. to time (t),

Or, 2x.$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ + 2y.$\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = 0.

Or, 2 * 2$\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ + 2 * 1.5 * $\left( { - \frac{{12}}{{100}}} \right)$ = 0      $\left[ {\frac{{{\rm{dy}}}}{{{\rm{dt}}}} =  - \frac{{12}}{{100}}\frac{{\rm{m}}}{{{\rm{sec}}}}} \right]$

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{{36}}{{100{\rm{*}}4}}$ = $\frac{9}{{100}}$.

So, speed of the lower end $\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)$ = $\frac{9}{{100}}$ m./sec = 9cm/sec.