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We have learned that the change in distance per unit time is called speed. Similarly, the change in velocity per unit time is called acceleration. This change is also known as an Instantaneous change. In chapter 6, exercise 6.1, we will learn the concept of differential change. All these NCERT solutions given in Class 12 Maths chapter 6 exercise 6.1 provide a deeper knowledge of the concept of rate of change of quantities that is discussed in the NCERT book. These NCERT solutions are created by a subject matter expert at Careers360. These solutions will give you concept clarification and help you to know the proper methods to solve the questions.
Question 1: a) Find the rate of change of the area of a circle with respect to its radius r when
Answer:
Given: Length $x$ of a rectangle is decreasing at the rate $\frac{dx}{dt} = -5\ \text{cm/min}$ (negative sign indicates decrease),
the width $y$ is increasing at the rate $\frac{dy}{dt} = 4\ \text{cm/min}$.
To find: $\frac{dP}{dt}$ at $x = 8\ \text{cm}$ and $y = 6\ \text{cm}$, where $P$ is the perimeter.
Perimeter of rectangle: $P = 2(x + y)$
$\frac{dP}{dt} = \frac{d}{dt}[2(x + y)] = 2\left( \frac{dx}{dt} + \frac{dy}{dt} \right) = 2(-5 + 4) = -2\ \text{cm/min}$
Hence, the perimeter is decreasing at the rate of $2\ \text{cm/min}$.
Question 1: b) Find the rate of change of the area of a circle with respect to its radius r when
r = 4 cm
Answer:
Area of the circle (A) = $\pi r^{2}$
Rate of change of the area of a circle with respect to its radius r = $\frac{dA}{dr}$ = $\frac{d(\pi r^{2})}{dr}$ = $2 \pi r$
So, when r = 4, Rate of change of the area of a circle = $2 \pi (4)$ = $8 \pi$
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is $8 \pi$
Answer:
The volume of the cube(V) = $x^{3}$ where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of $8\ \text{cm}^3/\text{s}$
we can write $\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ ( By chain rule)
$\frac{dV}{dt} = 8 = \frac{dV}{dx}.\frac{dx}{dt}$
$\frac{dx^{3}}{dx}.\frac{dx}{dt} = 8$ $\Rightarrow 3x^{2}.\frac{dx}{dt} = 8$
$\frac{dx}{dt} = \frac{8}{3x^{2}}$ - (i)
Now, we know that the surface area of the cube(A) is $6x^{2}$
$\frac{dA}{dt} = \frac{dA}{dx}.\frac{dx}{dt} = \frac{d6x^{2}}{dx}.\frac{dx}{dt} = 12x. \frac{dx}{dt}$ - (ii)
from equation (i) we know that $\frac{dx}{dt} = \frac{8}{3x^{2}}$
put this value in equation (i)
We get,
$\frac{dA}{dt} = 12x. \frac{8}{3x^{2}} = \frac{32}{x}$
It is given in the question that the value of edge length(x) = 12cm
So,
$\frac{dA}{dt} = \frac{32}{12} = \frac{8}{3} \ \text{cm}^2/\text{s}$
Answer:
Radius of a circle is increasing uniformly at the rate $\left ( \frac{dr}{dt} \right )$ = 3 cm/s
Area of circle(A) = $\pi r^{2}$
$\frac{dA}{dt} =\frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dA}{dt} =\frac{d \pi r^{2}}{dr}.\frac{dr}{dt} = 2\pi r \times 3 = 6\pi r$
It is given that the value of r = 10 cm
So,
$\frac{dA}{dt} = 6\pi \times 10 = 60\pi\ \text{cm}^2/\text{s}$
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is $60\pi\ \text{cm}^2/\text{s}$
Answer:
It is given that the rate at which edge of cube increase $\left ( \frac{dx}{dt} \right )$ = 3 cm/s
The volume of cube = $x^{3}$
$\frac{dV}{dt} = \frac{dV}{dx}.\frac{dx}{dt}$ (By chain rule)
$\frac{dV}{dt} = \frac{d{x^3}}{dx} \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt} = 3x^2 \times 3 = 9x^2\ \text{cm}^3/\text{s}$
It is given that the value of x is 10 cm
So,
$\frac{dV}{dt} = 9(10)^{2} = 9\times 100 = 900\ \text{cm}^3/\text{s}$
Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is $900\ \text{cm}^3/\text{s}$
Answer:
Given = $\frac{dr}{dt} = 5\ \text{cm/s}$
To find = $\frac{dA}{dt}$ at r = 8 cm
Area of the circle (A) = $\pi r^{2}$
$\frac{dA}{dt} = \frac{dA}{dr}.\frac{dr}{dt}$ (by chain rule)
$\frac{dA}{dt} = \frac{d(\pi r^2)}{dr} \cdot \frac{dr}{dt} = 2\pi r \times 5 = 10\pi r = 10\pi \times 8 = 80\pi \ \text{cm}^2/\text{s}$
Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is $80\pi \ \text{cm}^2/\text{s}$
Answer:
Given = $\frac{dr}{dt} = 0.7\ \text{cm/s}$
To find = $\frac{dC}{dt}$, where C is circumference
We know that the circumference of the circle (C) = $2\pi r$
$\frac{dC}{dt} = \frac{dC}{dr} \cdot \frac{dr}{dt}$ (by chain rule)
$\frac{dC}{dt} = \frac{d(2\pi r)}{dr} \cdot \frac{dr}{dt} = 2\pi \times 0.7 = 1.4\pi\ \text{cm/s}$
Hence, the rate of increase of its circumference is $1.4\pi\ \text{cm/s}$
Answer:
Given: Length $x$ of a rectangle is decreasing at the rate $\frac{dx}{dt} = -5$ cm/min (negative sign indicates decrease),
the width $y$ is increasing at the rate $\frac{dy}{dt} = 4$ cm/min.
To find: $\frac{dP}{dt}$ at $x = 8$ cm and $y = 6$ cm, where $P$ is the perimeter.
Perimeter of rectangle: $P = 2(x + y)$
$\frac{dP}{dt} = \frac{d}{dt}[2(x + y)]$
$= 2\left( \frac{dx}{dt} + \frac{dy}{dt} \right)$
$= 2(-5 + 4)$
$= -2\ \text{cm/min}$
Hence, the perimeter decreases at the rate of $2$ cm/min.
Answer:
Given = Length x of a rectangle is decreasing at the rate $(\frac{dx}{dt})$ = -5 cm/minute (−ve sign indicates decrease in rate)
The width y is increasing at the rate $(\frac{dy}{dt})$ = 4 cm/minute
Area of rectangle = $xy$
$\frac{dA}{dt} = \frac{d(xy)}{dt} = \left( x\frac{dy}{dt} + y\frac{dx}{dt} \right) = \left( 8 \times 4 + 6 \times (-5) \right) = (32 - 30) = 2\ \text{cm}^2/\text{minute}$
Hence, the rate of change of area is $2\ \text{cm}^2/\text{minute}$
Answer:
Given = $\frac{dV}{dt} = 900\ \text{cm}^3/\text{s}$
To find = $\frac{dr}{dt}$ at $r = 15$ cm
Volume of sphere (V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = \frac{d\left(\frac{4}{3}\pi r^{3}\right)}{dr} \cdot \frac{dr}{dt} = \frac{4}{3}\pi \times 3r^{2} \cdot \frac{dr}{dt}$
$\frac{dV}{dt} = 4\pi r^{2} \cdot \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^{2}} = \frac{900}{4\pi \times (15)^2} = \frac{900}{900\pi} = \frac{1}{\pi}\ \text{cm}/\text{s}$
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is $\frac{1}{\pi}\ \text{cm}/\text{s}$
Answer:
We need to find the value of $\frac{dV}{dr}$ at $r = 10$ cm.
The volume of the sphere (V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dr} = \frac{d\left(\frac{4}{3}\pi r^{3}\right)}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi\ \text{cm}^3/\text{s}$
Hence, the rate at which its volume is increasing with the radius when the latter is 10 cm is $400\pi\ \text{cm}^3/\text{s}$
Answer:
Let $h$ be the height of the ladder and $x$ be the distance of the foot of the ladder from the wall.
It is given that $\frac{dx}{dt} = 2\ \text{cm/s}$.
We need to find the rate at which the height of the ladder decreases $\left(\frac{dh}{dt}\right)$.
Length of ladder ($L$) = 5 m and $x = 4$ m (given).
By Pythagoras theorem, we can say that
$h^{2} + x^{2} = L^{2}$
$h^{2} = L^{2} - x^{2}$
$h = \sqrt{L^{2} - x^{2}}$
Differentiate on both sides with respect to $t$:
$\frac{dh}{dt} = \frac{d(\sqrt{L^{2} - x^{2}})}{dx} \cdot \frac{dx}{dt} = \frac{1}{2} \cdot \frac{-2x}{\sqrt{5^{2} - x^{2}}} \cdot \frac{dx}{dt} = \frac{-x}{\sqrt{25 - x^{2}}} \cdot \frac{dx}{dt}$
At $x = 4$:
$\frac{dh}{dt} = \frac{-4}{\sqrt{25 - 16}} \cdot 2 = \frac{-4}{3} \cdot 2 = \frac{-8}{3}\ \text{cm/s}$
Hence, the rate at which the height of the ladder decreases is $\frac{8}{3}\ \text{cm/s}$.
Answer:
We need to find the point at which $\frac{dy}{dt} = 8\frac{dx}{dt}$
Given the equation of curve = $6y = x^3 + 2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0$
$= 3x^{2}.\frac{dx}{dt}$
$\frac{dy}{dt} = 8\frac{dx}{dt}$ (required condition)
$6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}$
$3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}$ $\Rightarrow x^{2} = \frac{48}{3} = 16$
$x = \pm 4$
when x = 4 , $y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11$
and
when x = -4 , $y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3}$
So , the coordinates are
$(4,11)$ and $(-4,\frac{-31}{3})$
Answer:
It is given that $\frac{dr}{dt} = \frac{1}{2} \ \text{cm/s}$
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ \text{cm}^{3}/\text{s}$
Hence, the rate of change in volume is $2\pi \ \text{cm}^{3}/\text{s}$
Answer:
Volume of sphere (V) = $\frac{4}{3}\pi r^{3}$
Diameter = $\frac{3}{2}(2x+1)$
So, radius (r) = $\frac{3}{4}(2x+1)$
$\frac{dV}{dx} = \frac{d(\frac{4}{3}\pi r^{3})}{dx} = \frac{d\left(\frac{4}{3}\pi \left(\frac{3}{4}(2x+1)\right)^{3}\right)}{dx} = \frac{4}{3}\pi \times 3 \times \frac{27}{64}(2x+1)^{2} \times 2$
$= \frac{27}{8}\pi (2x+1)^{2}$
Answer:
Given = $\frac{dV}{dt} = 12\ \text{cm}^{3}/\text{s}$ and $h = \frac{1}{6}r$
To find = $\frac{dh}{dt}$ at $h = 4\ \text{cm}$
Solution:-
Volume of cone (V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = \frac{d\left(\frac{1}{3}\pi (6h)^{2}h\right)}{dh} \cdot \frac{dh}{dt} = \frac{1}{3}\pi \times 36 \times 3h^{2} \cdot \frac{dh}{dt}$
$= 36\pi \times (4)^{2} \cdot \frac{dh}{dt}$
$\frac{dV}{dt} = 576\pi \cdot \frac{dh}{dt}$
Find the marginal cost when 17 units are produced.
Answer:
Marginal cost (MC) = $\frac{dC}{dx}$
$C ( x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000$
$\frac{dC}{dx} = \frac{d(.007x^{3} - .003x^{2}+15x+400)}{dx} = 3\times .007x^{2} - 2\times.003x+15$
$= .021x^{2} - .006x + 15$
Now, at x = 17
MC $= .021(17)^{2} - .006(17) + 15$
$= 6.069 - .102 + 15$
$= 20.967$
Hence, marginal cost when 17 units are produced is 20.967
Question 16: The total revenue in Rupees received from the sale of x units of a product is
given by $R ( x) = 13 x^2 + 26 x + 15$
Find the marginal revenue when x = 7
Answer:
Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 13 x^2 + 26 x + 15$
$\frac{dR}{dx} = \frac{d(13x^{2}+26x+15)}{dx} = 13\times2x+ 26 = 26(x+1)$
at x = 7
$\frac{dR}{dx} = 26(7+1) = 26\times8 = 208$
Hence, marginal revenue when x = 7 is 208
Question 17: The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) $10\pi$
(B) $12\pi$
(C) $8\pi$
(D) $11\pi$
Answer:
Area of circle (A) = $\pi r^{2}$
$\frac{dA}{dr} = \frac{d(\pi r^{2})}{dr} = 2\pi r$
Now, at $r = 6\ \text{cm}$
$\frac{dA}{dr} = 2\pi \times 6 = 12\pi\ \text{cm}^{2}/\text{s}$
Hence, the rate of change of the area of a circle with respect to its radius $r$ at $r = 6\ \text{cm}$ is $12\pi\ \text{cm}^{2}/\text{s}$
Hence, the correct answer is $12\pi$
(A) $116$
(B) $96$
(C) $90$
(D) $126$
Answer:
Marginal revenue = $\frac{dR}{dx}$
$R ( x) = 3 x^2 + 36 x + 5$
$\frac{dR}{dx} = \frac{d(3x^{2}+36x+5)}{dx} = 3\times2x+ 36 = 6(x+6)$
at x = 15
$\frac{dR}{dx} = 6(15+6) = 6\times21 = 126$
Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is 126
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NCERT Solutions Subject Wise
Here are links to NCERT textbook solutions for other subjects. Students can explore and review these organised solutions to gain a better understanding.
Students may refer to these NCERT exemplar links for additional practice purposes.
Including miscellaneous, there are 6 exercises in the unit Applications of Derivatives
Two multiple-choice questions are discussed in the NCERT exercise 6.1
There are six solved examples
dy/dx means the rate of change of y with respect to x
dy/dx is positive if y increases as x increases
The topic number 6.3 increasing and decreasing function is coming after exercise 6.1.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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