Ex 6.1 class 12 maths ncert solutions

(a) We know that area of a circle (A) is \( \mathrm{A}=\pi \mathrm{r}^{2} \)
Then, Rate of change of the area with respect to its radius is given by, \( \frac{d A}{d r}=\frac{d\left(\pi r^{2}\right)}{d r}=2 \pi \mathrm{r} \)
When \( \mathrm{r}=3 \mathrm{~cm} \)
\(\frac{d A}{d r}=2 \pi(3)=6 \pi
\)
Then
Therefore, the area of the circle is changing at the rate of \( 6 \pi \mathrm{cm}^2 / \mathrm{s} \) when its radius is 3 cm .

(b) We know that area of a circle (A) is \( \mathrm{A}=\pi \mathrm{r}^{2} \)
Then, Rate of change of the area with respect to its radius is given by, \( \frac{d A}{d r}=\frac{d\left(\pi r^{2}\right)}{d r}=2 \pi \mathrm{r} \)
When \( \mathrm{r}=4 \mathrm{~cm} \)
Then
\(
\frac{d A}{d r}=2 \pi(4)=8 \pi
\)
Therefore, the area of the circle is changing at the rate of \( 8 \pi \mathrm{cm}^{2} / \mathrm{s} \) when its radius is 4 cm .

Let a be the length of a side, \( V \) be the volume and \( S \) be the surface area of the cube.
Then, \( V=a^{3} \) and \( S=6 a^{2} \) where \( a \) is the function of time \( t \).
Now, It is given that \( \frac{d V}{d t}=8 \mathrm{~cm}^{3} / \mathrm{s} \)
Then, by using the chain rule, we get,
\( 8=\frac{d V}{d t}=\frac{d\left(a^{3}\right)}{d t}=\frac{d\left(a^{3}\right)}{d t} \cdot \frac{d a}{d t}=3 a^{2} \cdot \frac{d a}{d t} \)
\( \Rightarrow \frac{d a}{d t}=\frac{8}{3 a^{2}} \ldots(i)\)
\( \frac{d s}{d t}=\frac{d\left(6 a^{2}\right)}{d t}=\frac{d\left(6 a^{2}\right)}{d t} \cdot \frac{d a}{d t}=12 \mathrm{a} \cdot \frac{d a}{d t}=12 a. \frac{8}{3 a^{2}}=\frac{32}{a} \ldots(ii)\)
So, when \( \mathrm{a}=12 \mathrm{~cm} \), then, \( \frac{d s}{d t}=\frac{32}{12}=\frac{8}{3} \mathrm{~cm}^{3} / \mathrm{s} \)
Therefore, if the length of the edge of the cube is 12 cm , then the surface area is increasing at the rate of \( \frac{8}{3} \mathrm{~cm}^{3} / \mathrm{s} \).

We know that area of a circle (A) is \( \mathrm{A}=\pi \mathrm{r}^{2} \).
Then, Rate of change of the area with respect to time is given by, \( \frac{d A}{d t}=\frac{d\left(\pi r^{2}\right)}{d t}=\frac{d\left(\pi r^{2}\right)}{d t} \cdot \frac{d r}{d t}=2 \pi \mathrm{r} \cdot \frac{d r}{d t} \)
It is given that
\(\frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{s}\)
Then, \( \frac{d A}{d r}=2 \pi(3)=6 \pi \mathrm{r} \)
Thus, when \( \mathrm{r}=10 \mathrm{~cm} \)
Then, \( \frac{d A}{d r}=6 \pi(10)=60 \pi \mathrm{cm}^{2} / \mathrm{s} \).
Therefore, the rate at which the area of the circle is increasing when the radius is 10 cm is \( 60 \pi \mathrm{cm}^{2} / \mathrm{s} \).

Let \(x\) be the length of a side and V be the volume of the cube, then
\( \mathrm{v}=x^{3} \)
\( \frac{d v}{d t}=\frac{d\left(x^{3}\right)}{d t}=3 x^{2} \cdot \frac{d x}{d t}=3 x^{2} \cdot \frac{d x}{d t} \ldots \).by chain rule
It is given that
\(
\frac{d x}{d t}=3 \mathrm{~cm} / \mathrm{s}
\)
Then, \( \frac{d v}{d r}=3 x^{2}(3)=9 x^{2} \)
Thus, when \( \mathrm{x}=10 \mathrm{~cm} \)
Then, \( \frac{d v}{d r}=9(10)^{2}=900 \mathrm{~cm}^{3} / \mathrm{s} \).
Therefore, the volume of the cube increasing when the edge is 10 cm long is \( 900 \mathrm{~cm}^{3} / \mathrm{s} \).

We know that area of a circle (A) is \( \mathrm{A}=\pi \mathrm{r}^{2} \)
Then, Rate of change of the area with respect to time \( (\mathrm{t}) \) is given by, \( \frac{d A}{d t}=\frac{d\left(\pi r^{2}\right)}{d t}=\frac{d\left(\pi r^{2}\right)}{d t} \cdot \frac{d r}{d t}=2 \pi r \cdot \frac{d r}{d t} \ldots \).by chain rule
It is given that
\(
\frac{d r}{d t}=5 \mathrm{~cm} / \mathrm{s}
\)
Then, \( \frac{d A}{d r}=2 \pi(5)=10 \pi \mathrm{r} \)
Thus, when \( \mathrm{r}=8 \mathrm{~cm} \)
Then, \( \frac{d A}{d r}=10 \pi(8)=80 \pi \mathrm{cm}^{2} / \mathrm{s} \).
Therefore, when the radius of the circular wave is 8 cm , how fast is the enclosed area increasing is \( 80 \pi \mathrm{cm}^{2} / \mathrm{s} \).

We know that circumference of a circle ( C ) is \( \mathrm{C}=2 \pi \mathrm{r} \)
Then, Rate of change of circumference with respect to time \( (t) \) is given by,
\( \frac{d C}{d t}=\frac{d(2 \pi \mathrm{r})}{d t}=\frac{d(2 \pi \mathrm{r})}{d t} \cdot \frac{d r}{d t}=2 \pi \cdot \frac{d r}{d t} \)...by chain rule
It is given that
\(
\frac{d r}{d t}=0.7 \mathrm{~cm} / \mathrm{s}
\)
Then, \( \frac{d C}{d r}=2 \pi(0.7)=1.4 \pi \mathrm{cm} / \mathrm{s} \)
Therefore, the rate of increase of its circumference is \( 1.4 \pi \mathrm{cm} / \mathrm{s} \).

(a) It is given that the length \(( x )\) is decreasing at the rate of \( 5 \mathrm{~cm} / \) minute and the width \((y)\) is increasing at the rate of \( 4 \mathrm{~cm} / \) minute, then we have, \( \frac{d x}{d t}=-5 \mathrm{~cm} / \) minute and \( \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min} \)
The perimeter \( (\mathrm{P}) \) of a rectangle is given by:
\(P=2(x+y)
\)
\(\therefore \frac{d p}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
\(=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}
\)
Therefore, the rate of change of the perimeter is \( -2 \mathrm{~cm} / \mathrm{min} \).
(b) It is given that the length \(( x )\) is decreasing at the rate of \( 5 \mathrm{~cm} / \) minute and the width \(( y )\) is increasing at the rate of \( 4 \mathrm{~cm} / \) minute, then we have, \( \frac{d x}{d t}=-5 \mathrm{~cm} / \) minute and \( \frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min} \)
Now, the area (A) of the rectangle is given by
\(
\mathrm{A}=x \times y\)
\(\therefore \frac{d A}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}\)
\(=-5 y+4 x
\)
So, when \(x=8 \mathrm{~cm} \) and \( y=6 \mathrm{~cm} \), then,
\(
\frac{d A}{d t}=\frac{(-5 \times 6+4 \times 8) \mathrm{cm}^{2}}{\min }=2 \mathrm{~cm}^{2} / \mathrm{min}
\)
Therefore, the rate of change of the area of the rectangle is \( 2 \mathrm{~cm}^{2} / \mathrm{min} \).

Let V be the volume of the sphere, then
\(
\mathrm{V}=\frac{4}{3} \pi r^{3}
\)
Therefore, Rate of change of volume \( (\mathrm{V}) \) with respect to time \( (\mathrm{t}) \) is given by,
\( \frac{d v}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t} \cdot \frac{d r}{d t}=4 \pi r^{2} \cdot \frac{d r}{d t} \ldots \ldots \) by chain rule
It is given that
\(\frac{d v}{d t}=900 \mathrm{~cm}^{3} / \mathrm{s}\)
Then, \( 900=4 \pi r^{2} \cdot \frac{d r}{d t} \)
\( \Rightarrow \) when \( \mathrm{r}=15 \mathrm{~cm} \)
Then, \( \frac{d v}{d r}=\frac{225}{\pi 15^{2}}=\frac{1}{\pi} \)
Therefore, the rate at which the radius of the balloon increases when the radius is 15 cm is \( \frac{1}{\pi} \mathrm{cm} / \mathrm{s} \).

Let V be the volume of the sphere, then
\(\mathrm{V}=\frac{4}{3} \pi r^{3}\)
Therefore, Rate of change of volume \( (\mathrm{V}) \) with respect to its radius ( r ) is given by,
\( \frac{d V}{d r}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}=\frac{4}{3} \pi\left(3 r^{2}\right)=4 \pi r^{2} \ldots . \). by chain rule
Thus, when \( \mathrm{r}=10 \mathrm{~cm} \)
Then, \( \frac{d V}{d r}=4 \pi(10)^{2}=400 \pi \)
Therefore, the rate at which its volume is increasing with the radius when the later is 10 cm is \( 400 \pi \mathrm{cm}^{3} / \mathrm{s} \).

Let \( y \) be the height of the wall at which the ladder touches. Also, let the foot of the ladder be \(x\) m away from the wall.
we know that, by Pythagoras theorem,
\(
\Rightarrow x^{2}+y^{2}=25\)
\(\Rightarrow y=\sqrt{25-x^{2}}
\)
Then, the rate of change of height \((y)\) with respect to time \((t)\) is given by: \( \frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t} \)
Now, it is given that \( \frac{d x}{d t}=\frac{2 c m}{s} \).
Therefore,
\(\frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}\)
And when \(x=4 \mathrm{~m} \), then
\(\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=\frac{8}{3}\)
Therefore, the height of the ladder on the wall is decreasing at the rate of \( \frac{8}{3} \mathrm{~m} / \mathrm{s} \).

It is given that equation of the curve is \( 6 y=x^{3}+2 \).
The rate of change of the position of the particle with respect to time \( (t) \) is given by:
\(
6 \frac{d y}{d t}=3 x^{2} \cdot \frac{d x}{d t}+0\)
\(\Rightarrow 2 \frac{d y}{d t}=x^{2} \cdot \frac{d x}{d t} \ldots(i)\)
When the \(y\)- coordinate of the particle changes 8 times as fast as the \( x- \) coordinate then, we have,
\(\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right)\)
So, putting the value in (1), we get,
\(
\Rightarrow 2\left(8 \frac{d x}{d t}\right)=x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow 16 \frac{d x}{d t}=x^{2} \cdot \frac{d x}{d t}\)
\(\Rightarrow\left(x^{2}-16\right) \cdot \frac{d x}{d t}=0\)
\(\Rightarrow x^{2}=16\)
\(\Rightarrow x= \pm 4
\)
So, when \( x= \pm 4 \), then \( y=\frac{(4)^{3}+2}{6}=\frac{66}{6}=11 \)
And
So, when \( x=-4 \), then, \( y=\frac{-62}{6}=\frac{-31}{2} \)
Therefore, the points required on the curve are \( (4,11) \) are \( (-4,\frac{-31}{2}) \).

As, the air bubble is in the shape of a sphere.
Therefore, V be the volume of the sphere, then \( \mathrm{V}=\frac{4}{3} \pi r^{3} \)
Therefore, Rate of change of volume \( (\mathrm{V}) \) with respect to time \( (\mathrm{t}) \) is given by,
\( \frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t} \cdot \frac{d r}{d t}=4 \pi r^{2} \cdot \frac{d r}{d t} \ldots \)by chain rule
It is given that
\( \frac{d r}{d t}=\frac{1}{2} \mathrm{~cm} / \mathrm{s} \)
when \( \mathrm{r}=1 \mathrm{~cm} \)
Then, \( \frac{d V}{d r}=4 \pi 1^{2} \cdot \frac{1}{2}=2 \pi c \mathrm{~m}^{3} / \mathrm{s} \)
Therefore, the rate is the volume of the bubble increasing when the radius is 1 cm is \( 2 \pi \mathrm{cm}^{3} / \mathrm{s} \).

Let V be the volume of the sphere, then
\(
\mathrm{V}=\frac{4}{3} \pi r^{3}
\)
It is given that, Diameter \( =\frac{3}{2}(2 x+1) \)
\(
\Rightarrow \mathrm{r}=\frac{3}{4}(2 x+1)\)
\(\therefore \mathrm{V}=\frac{4}{3} \pi r\left(\frac{3}{4}\right)^{3}(2 x+1)^{3}\)
\(\Rightarrow \mathrm{V}=\frac{9}{16} \pi(2 x+1)^{3}
\)
Now, the rate of change of its volume with respect to \(x\) is as
\(
\frac{d V}{d t}=\frac{9}{16} \pi(2 x+1)^{3}=\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2\)
\(=\frac{27}{8} \pi(2 x+1)^{2}
\)
Therefore, the rate of change of its volume with respect to \(x\) is \( \frac{27}{8} \) \( \pi(2 x+1)^{2} \mathrm{~cm}^{3} / \mathrm{s} \).

Let V be the volume of the cone, then
\(
\mathrm{V}=\frac{1}{3} \pi r^{2} h
\)
It is given that, \( \mathrm{h}=\frac{3}{2} \mathrm{r} \)
\(
\Rightarrow r=6 h
\)
Therefore, \( \mathrm{V}=\frac{1}{3} \pi(6 h)^{2} h=12 \pi h^{3} \)
Now, the rate of change of its volume with respect to time \( (\mathrm{t}) \) is given by:
Now, the rate of change of its volume with respect to \(x\) is as
\( \frac{d v}{d t}=12 \pi \frac{d(h)^{2}}{d t} \ldots\) by chain rule
\(
\Rightarrow \frac{d v}{d t}=12 \pi\left(3 h^{2}\right) \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t}\ldots(i)\)
Now, it is also given that
\( \frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s} \) and \( \mathrm{h}=4 \mathrm{~cm} \)
So, putting the value in equation (1), we get
\(
\Rightarrow 12=36 \pi 4^{2} \frac{d h}{d t}\)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi(16)}=\frac{1}{48 \pi}
\)
Therefore, the height of the sand cone increasing when the height is 4 cm is \( \frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \).

Marginal cost (MC) is the rate of change of total cost with respect to output.
Then,
\(
\mathrm{MC}=\frac{d C}{d x}=0.007\left(3 x^{2}\right)-0.003(2 x)+15\)
\(=0.021 x^{2}-0.006 x+15
\)
So, when \( x=17 \) then,
\(
\mathrm{MC}=0.021\left(17^{2}\right)-0.006(17)+15\)
\(=0.021(289)-0.102+15\)
\(=20.967
\)
Therefore, the marginal cost when \(17\) units are produced is Rs. \(20.967\).

Marginal revenue (MR) is the rate of change of total revenue with respect to the number of units sold.
Then,
\(
\mathrm{MR}=\frac{d R}{d x}=13(2 x)+26=26 x+26
\)
So, when \( x=7 \) then,
\(
\mathrm{MR}=26(7)+26=208
\)
Therefore, the marginal revenue when \( x=7 \) is Rs. 208 .

We know that area of a circle (A) is \( \mathrm{A}=\pi \mathrm{r}^{2} \)
Then, Rate of change of the area with respect to its radius is given by, \( \frac{d A}{d r}=\frac{d\left(\pi r^{2}\right)}{d r}=2 \pi \mathrm{r} \)
When \( \mathrm{r}=6 \mathrm{~cm} \)
Then,\( \frac{d A}{d r}=2 \pi(6)=12 \pi \mathrm{r} \)
Therefore, the area of the circle is changing at the rate of \( 12 \pi \mathrm{cm}^2 / \mathrm{s} \) when its radius is 6 cm .

Marginal revenue (MR) is the rate of change of total revenue with respect to the number of units sold.
Then,
\(
\mathrm{MR}=\frac{d R}{d x}=3(2 x)+36=6 x+36
\)
So, when \( x=15 \) then,
\(
M R=6(15)+36=126
\)
Therefore, the marginal revenue when \( x=15 \) is Rs. 126 .