ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius when the radius is 6 cm. (NCERT)
Solution:
Let r be the radius of circle
Then area of circle = A = πr²
∴ \(\frac{d A}{d r}\) = 2πr
When r = 6 cm ;
\(\frac{d A}{d r}\) = 2 π × 6 cm
= 12 π cm
Thus, the area of the circle changing at the rate of 12 π cm²/cm.

Question 2.
If the radius of a circle is increasing at the rate of 0.5 cm/sec, at what rate is its circumference increasing?
Solution:
Let r be the radius of circle at any time t
and let S be the circumference of circle.
Then S = 2πr
∴ \(\frac{d S}{d t}\) = 2π \(\frac{d r}{d t}\)
Given \(\frac{d r}{d t}\) = 0.5 cm/sec
Thus, \(\frac{d S}{d t}\) = 2π × 0.5 = π cm/sec
Hence, the rate at which circumference of circle is increasing be π cm/sec.

Question 2 (old).
If the radius of a circle is increasing at the rate of 0.7 cm/sec, at what rate is its circumference increasing ?(NCERT)
Solution:
Let r be the radius of circle then
\(\frac{d r}{d t}\) = 0.7 cm/sec.
Let c = circumference of circle = 2πr
∴ \(\frac{d c}{d t}\) = 2π × \(\frac{d r}{d t}\)
= 2π × 0.7
= 1.4π cm/sec.

Question 3.
If the radius of a circle is increasing at the rate of 3 cm/sec, at what rate is its area increasing when its radius is 2 cm?
Solution:
Let r be the radius of circle at any time t
Let A = area of circle = πr²
Given \(\frac{d r}{d t}\) = 3cm/sec;
r = 2 cm
∴ \(\frac{d A}{d t}\) = 2πr = 2π × 2 × 3
= 12π cm²/sec.

Question 3(old).
If the radius of a circle is increasing at the rate of 3 cm/sec, at what rate is its area increasing when its radius is 10 cm ? (NCERT)
Solution:
Let A be the area of circle with radius r
Then A = πr²
On differentiating w.r.t. t, we have
\(\frac{d A}{d t}\) = 2πr \(\frac{d r}{d t}\) ………….(1)
Given, \(\frac{d r}{d t}\) = 3 cm/sec
and r = 10 cm
∴ from (1) ;
\(\frac{d A}{d t}\) = (2π × 10 × 3) cm² / sec
Thus, the required area of circle is increasing at the rate of 60π cm²/sec.

Question 4.
If the sides of a square are decreasing at the rate of 1.5 cm/sec, at what rate is its perimeter decreasing ?
Solution:
Let P be the perimeter of square with side x cm.
Then P = 4x
Diff. both sides w.r.t. t, we have
\(\frac{d p}{d t}\) = 4 \(\frac{d x}{d t}\) …………(1)
Given, \(\frac{d r}{d t}\) = – 1.5 cm/sec.
∴ from (1) ;
\(\frac{d p}{d t}\) = – (4 × 1.5) cm/sec = – 6 cm/sec.
Hence the perimeter of square is decreasing at the rate of 6 cm/sec.

Question 5.
If an edge of a variable cube is increasing at the rate of 10 cmlsec, at what rate is its volume increasing when its edge is 5 cm?
Solution:
Let x be the edge of variable cube at any time t
Let V = volume of cube = x³
Given, \(\frac{d x}{d t}\) = 10 cm / sec ;
x = 5 cm
∴ \(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\)
= 3 × 5² × 10 cm³ / sec
= 750 cm³ / sec

Question 5 (old).
If an edge of a variable cube is increasing at the rate of 3 cm/sec, at what rate is its volume increasing when its edge is 10 cm ? (NCERT)
Solution:
Let x be the length of edge of cube s.t. \(\frac{d x}{d t}\) = 3 cm/sec.
Then V = volume of cube = x³
∴ \(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\)
Given, x = 10 cm
∴ \(\frac{d V}{d t}\) = 3 × (10)² × 3 = 900 cm³/sec.

Question 6.
A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with respect to radius when the radius is 10 cm. (NCERT)
Solution:
Let V be the volume of spherical balloon with radius r.
Then V = volume of spherical baloon = \(\frac{4}{3}\) πr³
Differentiating both sides w.r.t. r. we have
\(\frac{d V}{d r}\) = \(\frac{4 \pi}{3}\) × 3r² = 4πr²
When r = 10 cm ;
\(\frac{d V}{d r}\) = 4π × 10² cm² / cm
= 400 π cm³ / cm
Hence, the volume of spherical balloon is increasing at the rate of 400 cm³/cm.

Question 7.
If the radius of a balloon which always remains spherical is increasing at the rate of 1.5 cm/sec, at what rate is ¡t surface area increasing when its radius is 12 cm?
Solution:
Let r be the radius of spherical balloon at any time t
and let S be the surface area of spherical balloon.
Then S = 4πr²
⇒ \(\frac{d S}{d t}\) = 8πr \(\frac{d r}{d t}\) …………….(1)
Given r = 12 cm;
\(\frac{d r}{d t}\) = 1.5 cm/sec
from (1);
\(\frac{d S}{d t}\) = 8π × 12 × 15 cm² /sec
= 144 π cm²/sec.
Thus the surface area of spherical balloon is increasing at the rate 144 π cm²/sec.

Question 8.
If the radius of a soap bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec, at what rate is its volume increasing when the radius is 1 cm ? (NCERT)
Solution:
Let r be the radius of bubble and V be the volume of air bubble
Then V = \(\frac{4}{3}\) πr³
and \(\frac{d r}{d t}\) = 0.5 cm/sec.
Diff. both sides w.r.t. t; we have
∴ \(\frac{d V}{d t}\) = \(\frac{4 \pi}{3} \times 3 r^2 \frac{d r}{d t}\)
= 4πr² \(\frac{d r}{d t}\)
When r = 1 cm
∴ \(\frac{d V}{d t}\) = (4π × 1² × 0.5) cm³/sec
= 2π cm³/sec.

Question 9.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Find the rate at which the depth of the wheat is increasing, take π = 3.14. (NCERT)
Solution:
Let V be the volume of cylindrical tank with radius r.
Then V = πr²h ……….(1)
where h be the height of cylindrical tank.
On differentiating eqn. (1) w.r.t. t, we get
\(\frac{d \mathrm{~V}}{d t}=\pi r^2 \frac{d h}{d t}\) …………..(2)
Given \(\frac{d V}{d t}\) = 314 m³/hr ;
r = 10 cm
∴ from (2) ;
314 = 3.14 × (10)³
⇒ \(\frac{d h}{d t}\) = \(\frac{314}{314}\) = 1 m/hr
Thus, the required rate at which the depth of wheat is increasing be 1 m/hr.

Question 10.
(i) The total revenue in rupees received from the sale of x units of a product is given by
R (x) = 3x² + 36x + 5.
Find the marginal revenue when x = 5. (NCERT)
(ii) The total cost C (x) associated with the production of x units of an item is given by C (x) = 0.005x³ – 0.02x³ + 30x + 5000. Find the marginal cost when 3 units are produced.
Solution:
(i) Given R (x) = 3x² + 3 6x + 5
We know that marginal revenue is the rate of change of total revenue w.r.t. to number of units sold.
∴ \(\frac{d R}{d t}\) = 6x + 36
When x = 5 then,
\(\frac{d R}{d t}\) = 30 + 36 = Rs. 66

(ii) The total cost C (x) associated with the production of x unitsof an item is given by
C (x) = 0.005 x³ – 0.02 x² + 30x + 5000
∴ Marginal cost = \(\frac{d}{d x}\) C(x)
= 0.005 × 3x² – 0.02 × 2x + 30
= 0 015 x² – 0.04x + 30
∴ required marginal cost when 3 units are produced = [C’ (x)]_{x = 3}
= 0.015 × 3² – 0.04 × 3 + 30
= 0.135 – 0.12 + 30
= Rs. 30.015.

Question 11.
Find the rate of change of area of a circle of radius r with regard to its radius. How V\ fast is the area changing with respect to the radius when the radius is 5 cm ? (NCERT)
Solution:
Let A be the area of circle of radius r
Then A = π r² ;
differentiating both sides w.r.t. r, we have
\(\frac{d A}{d r}\) = 2πr
where r = 5 cm
∴ \(\frac{d A}{d r}\) = (2 π × 5) cm²/cm
= 10π cm²/cm

Question 12.
Find the rate of change of the whole surface of a closed circular cylinder of radius r and height h with respect to change in radius.
Solution:
Let S be the total surface area of closed cylinder with radius r and height h.
Then S = 2πr² + 2πrh
= 2π (2r + h) unit² / unit
On differentiating eqn. (1) w.r.t. r, we have
\(\frac{d S}{d r}\) = 4πr + 2πh
= 2π (2 r + h) unit² /unit.
Hence, the required rate of change of surface area of cylinder w.r.t. radius be 2π (2r + h) units² /units.

Question 13.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/sec. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing ? (NCERT)
Solution:
Let r be the radius of circular wave, then area of circular wave = A = πr²
given \(\frac{d r}{d t}\) = 5 cm/sec
and r = 8 cm
∴ \(\frac{d A}{d t}\) = 2π × r × \(\frac{d r}{d t}\)
= 2π × 8 × 5
= 80 π cm²/sec

Question 14.
A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing ? (NCERT)
Solution:
Let r the radius of circular wave, then area of circular wave = A = πr²
given \(\frac{d r}{d t}\) = 5 cm/sec
and r = 8 cm
∴ \(\frac{d A}{d t}\) = 2π × r × \(\frac{d r}{d t}\)
= 2π × 8 × 5
= 80π cm²/sec

Question 15.
The length x of a rectangle is decreasing at the rateof 5 cm/minute and the width y is increasing at the rate of 4 cm/ minute. When x = 8 cm and y = 6 cm, find the rate of change of
(i) the perimeter
(ii) the area of the rectangle.
Solution:
(i) Perimeter of rectangle = 2 (x + y) = p
∴ \(\frac{d p}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
Given \(\frac{d x}{d t}\) =- 5 cm/min
\(\frac{d y}{d t}\) = 4 cm/min
∴ \(\frac{d p}{d t}\) = 2 (- 5 + 4) = – 2 cm/min.

(ii) Let A = area of rectangle = xy
Diff. both sides w.r.t. t;
\(\frac{d p}{d t}=\frac{d y}{d t}+y \frac{d x}{d t}\)
= 8 × 4 + 6 × (- 5)
= 32 – 30
= 2 cm²/min

Question 16.
The volume of a cube is increasing at the rate of 9 cubic centimeters per \ 1o second. How fast is the surface area increasing when the length of the edge is 10 centimeters ? (NCERT)
Solution:
Let x be the length of edge of cube
Then V = volume of cube = x³ ………….(1)
and S = surface area of cube = 6x² …….(2)
From (1) ;
\(\frac{d V}{d t}\) = 3x² \(\frac{d x}{d t}\) ;
Also \(\frac{d V}{d t}\) = 9 cm³/sec.
9 = 3x² \(\frac{d x}{d t}\); also x = 10 cm
⇒ \(\frac{9}{3 \times 10^2}=\frac{d x}{d t}\)
⇒ \(\frac{d x}{d t}=\frac{3}{100}\) cm / sec
From (2) ;
\(\frac{d S}{d t}\) = 12x \(\frac{d x}{d t}\)
∴ \(\frac{d S}{d t}\) = 12 × 10 × \(\frac{3}{100}\)
= \(\frac{360}{100}\) cm² /sec.
⇒ \(\frac{d S}{d t}\) = 3.6 cm²/sec
Hence the surface area of the cube is increasing at the rate of 3.6 cm²/sec.

Question 16 (old).
The top of a ladder 6 metres long is resting against a vertical wall. Suddenly, c the ladder begins to slide outwards. At the instant when the foot of the ladder is 4 metres from the wall, it is sliding away at the rate of 0.5 m/sec. How fast is the top sliding downwards at this moment ? how far is the foot from the wall when the foot and the top are moving at the same rate ?
Solution:
Let PQ be the wall and AB be the ladder such that
AB = 6 m,
let AQ = y m and
QB = x m
Here x = 4 m and \(\frac{d x}{d t}\) = 0.5 m/sec

In right angled ΔAQB, we have
x² + y² = 36 ………..(1)
∴ 2x \(\frac{d x}{d t}\) + 2y \(\frac{d y}{d t}\) = 0
⇒ \(\frac{d y}{d t}\) = – \(\frac{x}{y}\) \(\frac{d x}{d t}\)
at x = 4 ;
\(\frac{d y}{d t}\) = – \(\frac{4}{y} \times \frac{1}{2}\)
= – \(\frac{2}{y}\) m / sec ……….(2)
From (1) ;
y² = 36 – x²
= 36 – 4² = 20
⇒ y = 2√5 m
∴ From (2) ; we have
\(\frac{d y}{d t}=\frac{-2}{2 \sqrt{5}}\) m/sec dt
= – \(\frac{1}{\sqrt{5}}\) m/sec
So the ladder top is sliding at the rate of \(\frac{1}{\sqrt{5}}\) m/sec.
Now we want to find the value of x such that
\(\frac{d x}{d t}=-\frac{d y}{d t}\)
∴ \(\frac{d x}{d t}=\frac{x}{y} \frac{d x}{d t}\)
⇒ x = y
from (1) ;
x² + x² = 36
⇒ x² = 18
⇒ x = 3√2 m
When foot and top are moving at the same rate and foot of wall is 3√2 m away from the wall.

Question 17.
A particle moves along the curve y = \(\frac{2}{3}\) x³ + 1. Find the points on the curve at which the y-coordinate is changing twice as fast as the x- coordinate.
Solution:
Let P (x, y) be any point on given curve
y = \(\frac{2}{3}\) x³ ……………(1)
also it is given that,
\(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\) …………(2)
From (1);
\(\frac{d y}{d t}\) = \(\frac{2}{3}\) × 3x² \(\frac{d x}{d t}\)
= 2x² \(\frac{d x}{d t}\)
∴ From (2) ; we have
2 \(\frac{d x}{d t}\) = 2x² \(\frac{d x}{d t}\)
⇒ x² = ± 1
When x = 1
∴ from (1); we have
y = \(\frac{2}{3}\) + 1 = \(\frac{5}{3}\)
When x = – 1
∴ from (2); we have
y = – \(\frac{2}{3}\) + 1 = \(\frac{1}{3}\)
Hence the required points on given curve be (1, \(\frac{5}{3}\)) and (- 1, \(\frac{1}{3}\)).

Question 18.
A man of height 2 meters walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of shadow increases.
Solution:
Let PQ be the lamp post and MN be the position of man at any time t and x metres be the distance of man from lamp post and y meters be the length of its shadow.
Then PQ = 6 m ;
MN = 2 m;

In similar Δ’s MLN and PLQ
\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{\mathrm{LN}}{\mathrm{LQ}}\)
⇒ \(\frac{2}{6}=\frac{y}{y+x}\)
⇒ 3y = y + x
⇒ y = \(\frac{x}{2}\)
∴ \(\frac{d y}{d t}=\frac{1}{2} \frac{d x}{d t}\)
but \(\frac{d x}{d t}\) = 5 km/hr
⇒ \(\frac{d y}{d t}=\frac{5}{2}\) km/hr.

Question 19.
A girl of height 1.6 m walks at the rate of 50 metres per minute away from a lamp which is 4 m above the ground. mHow fast is the girl’s shadow lengthening?
Solution:
Let AB be the lamp post where A be the position of the lamp then AB 4 cm.
Let CD be the position of girl at any time t and x be the distance of girl from lamp post.
Let y be the length of the shadow of girl.

Thus AECD and AEBA are similar A’s.
In similar A’s ECD and EBA, we have
\(\frac{E C}{E B}=\frac{C D}{B A}\)
⇒ \(\frac{y}{x+y}=\frac{16}{4}\)
⇒ \(\frac{y}{x+y}=\frac{16}{40}=\frac{2}{5}\)
⇒ 5y = 2x + 2y
⇒ 3y = 2x …………………(1)
Diff. eqn. (1) both sides w.r.t. t, we have
3 \(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\)
But given, \(\frac{d x}{d t}\) \(\frac{d x}{d t}\) = 50 m/min
∴ \(\frac{d y}{d t}\) = (2 × \(\frac{50}{3}\)) m/min
= \(\frac{100}{3}\) m/min
Thus the length of girl’s shaddow is increasing at the rate of 33\(\frac{1}{3}\) m/min.

Question 20.
A kite is 120 in high and 130 m of string c is out. If the kite is moving horizontally y at the rate of 5.2 m/sec, find the rate at which the string is being paid out at that instant.
Solution:
Let at any time t, Q be the position of kite.
Then in right angled ΔQPR we have,
y² = x² + (120)² ……………(1)

∴ 2y \(\frac{d y}{d t}\) = 2x \(\frac{d x}{d t}\) …………(2)
given,
\(\frac{d x}{d t}\) = 5.2 m/sec
and y = 130
∴ from (1) ;
(130)² – (120)² = x²
⇒ x² = 16900 – 14400
⇒ x² = 2500
⇒ x = 50
∴ from (2) ;
\(\frac{d y}{d t}=\frac{x}{y} \frac{d x}{d t}\)
= \(\frac{50}{130}\) × 5.2
= \(\frac{260}{130}\)
= 2 m/sec
Thus the required rate at which the dtring is being paid out be 2 m/sec.

Question 21.
A circular cone, with semi-vertical angle 45°, is fixed with its axis vertical and its vertex downwards. Water is poured into the cone at the rate of 2 cm3 per second. Find the rate at which the depth of the water is increasing when the depth is 4 cm.
Solution:
Let V be the volume of right circular cone with radius r and height h.

Also tan α = tan 45° = \(\frac{r}{h}\)
⇒ 1 = \(\frac{r}{h}\)
⇒ h = r
Then V = \(\frac{1}{3}\) πr² h
= \(\frac{\pi}{3}\)h³
On differentiating both sides w.r.t. t, we have
\(\frac{d \mathrm{~V}}{d t}=\frac{\pi}{3} \times 3 h^2 \frac{d h}{d t}\)
= πh² \(\frac{d h}{d t}\) …………(1)
Given \(\frac{d V}{d t}\) = 2 cm³ / sec ; h = 4 cm
∴ from (1) ;
2 = π × 4² \(\frac{d h}{d t}\)
⇒ \(\frac{d h}{d t}\) = \(\frac{1}{\pi}\) cm / sec
Thus the required rate at which the depth of water is increasing be \(\frac{1}{8 \pi}\) cm/sec.

Question 22.
Sand is pouring from a pipe at the rate of 12 cm³/sec. The falling sand forms a ,, cone on the ground in such a way that the height of the cone is always one- sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm ?
Solution:
Let r be the radius and h be the height of cone such that h = \(\frac{r}{6}\)
⇒ r = 6 h cm
Let V be the volume of circular cone
Then, V = \(\frac{\pi}{3}\)r²h
On differentiating both sides w.r.t. t, we have
⇒ \(\frac{d V}{d t}\) = 36πh² × \(\frac{d h}{d t}\) …………..(1)
Given, \(\frac{d V}{d t}\) = 12cm³ / sec ;
h = 4 cm
∴ from (1) ; we have
12 = 36π × 4² × \(\frac{d h}{d t}\)
⇒ \(\frac{d h}{d t}\) = \(\frac{12}{36 \pi \times 16}\) cm / sec
⇒ \(\frac{d h}{d t}=\frac{1}{48 \pi}\) cm / sec
Thus, the height of sand cone is increasing at the rate of \(\frac{1}{48 \pi}\) cm/sec.

Question 23.
A conical vessel whose height is 4 metres and of base radius 2 metres is being filled with water at the rate of 0.75 cubic metres per minute. Find the rate at which the level of the water is rising when the depth of water is 1.5 metres.
Solution:
Here 2 m is the radius and 4 m is the height of inverted cone.
If V be the volume at any time t
∴ \(\frac{d V}{d t}\) = \(\frac{3}{4}\) m³/mt. ………………(1)

At that instant, let the water form a cone of height h m and radius r m.
∴ Δ’s OBE and ODC are similar
i.e. \(\frac{E B}{C D}=\frac{E O}{C O}\)
⇒ \(\frac{2}{r}=\frac{4}{h}\)
⇒ h = 2r
⇒ r = \(\frac{h}{2}\)
Now volume of inverted cone V = \(\frac{1}{3}\) πr²h
⇒ V = \(\frac{1}{3} \pi\left(\frac{h}{2}\right)^2\) × h
V = \(\frac{1}{12}\) πh³
∴ \(\frac{d \mathrm{~V}}{d t}=\frac{1}{4} \pi h^2 \frac{d h}{d t}\)
⇒ \(\frac{3}{4}=\frac{\pi}{4} h^2 \frac{d h}{d t}\) [using (1)]
∴ \(\frac{d h}{d t}=\frac{3}{\pi h^2}\)
When the water level rising about the base = h = 1.5 m
Thus \(\frac{d h}{d t}=\frac{3}{\pi} \times \frac{4}{9}\) m/min
= \(\frac{4}{3 \pi}\) m/min.

Question 24.
Water is dripping out at a steady rate of 1 cu cm/sec through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the semi-vertical angle of the cone is \(\frac{\pi}{6}\). (NCERT Exampler)
Solution:
Let l be the slant height of conical vessel at any time t.
Then
V = \(\frac{\pi}{3}\) r²h

where r = radius of cone
h = height of cone
Here h = l cos \(\frac{\pi}{6}\) and
r = l sin \(\frac{\pi}{6}\)
∴ from (1) ;
V = \(\frac{\pi}{3}\) (l sin \(\frac{\pi}{6}\))² × l cos \(\frac{\pi}{6}\)
⇒ V = \(\frac{\pi}{3} l^3 \times\left(\frac{1}{2}\right)^2 \times \frac{\sqrt{3}}{2}\)
= \(\frac{\pi l^3 \sqrt{3}}{24}\)
Diff. both sides w.r.t. t, we have
\(\frac{d \mathrm{~V}}{d t}=\frac{\pi \sqrt{3}}{24} \times 3 l^2 \frac{d l}{d t}\)
= \(\frac{\sqrt{3} \pi}{8} l^2 \frac{d l}{d t}\) ………….(2)
Given \(\frac{d}{d x}\) = – 1 cu cm/ sec
and l = 4 cm
∴ from (2) ; we have
– 1 = \(\frac{\sqrt{3} \pi}{8} \times 4^2 \frac{d l}{d t}\)
⇒ \(\frac{d l}{d t}=-\frac{1}{2 \sqrt{3} \pi}\) cm / sec
Hence slant height of tne cone be decreasing at the rate of \(\frac{1}{2 \sqrt{3} \pi}\) cm/sec.