Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 6 Ex 6.1 page covers all 18 questions, solved step-by-step — applying dy/dx and the Chain Rule to rate-of-change problems on circles, cubes, spheres, cones and marginal cost & revenue, exactly the way CBSE Application of Derivatives answers are marked.
A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r.
(a) At r = 3: 2\pi(3) = 6\pi.
(b) At r = 4: 2\pi(4) = 8\pi.
Let edge = x: V = x^3 \Rightarrow \dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt}.
At x = 12: 8 = 3(144)\dfrac{dx}{dt} \Rightarrow \dfrac{dx}{dt} = \dfrac{1}{54}.
Also S = 6x^2 \Rightarrow \dfrac{dS}{dt} = 12x\dfrac{dx}{dt} = 12(12)\left(\dfrac{1}{54}\right) = \dfrac{8}{3}.
A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} = 2\pi(10)(3) = 60\pi.
V = x^3 \Rightarrow \dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt} = 3(100)(3) = 900.
A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} = 2\pi(8)(5) = 80\pi.
C = 2\pi r \Rightarrow \dfrac{dC}{dt} = 2\pi\dfrac{dr}{dt} = 2\pi(0.7) = 1.4\pi.
Given \dfrac{dx}{dt} = -5,\ \dfrac{dy}{dt} = 4.
(a) P = 2(x+y) \Rightarrow \dfrac{dP}{dt} = 2(-5+4) = -2.
(b) A = xy \Rightarrow \dfrac{dA}{dt} = \dfrac{dx}{dt}\cdot y + x\cdot\dfrac{dy}{dt} = (-5)(6) + (8)(4) = 2.
V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}. At r = 15: 900 = 4\pi(225)\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt} = \dfrac{1}{\pi}.
V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr} = 4\pi r^2 = 4\pi(100) = 400\pi.
Let x = distance of foot from wall, y = height on wall. By Pythagoras: x^2 + y^2 = 25 ...(1).
Differentiating: 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{x}{y}\cdot\dfrac{dx}{dt}.
At x = 4\text{ m}, (1) gives y = 3\text{ m}. With \dfrac{dx}{dt} = 0.02\text{ m/s}: \dfrac{dy}{dt} = -\dfrac{4}{3}(0.02) = -\dfrac{8}{3}\text{ cm/s}.
Differentiating 6y = x^3+2: 6\dfrac{dy}{dt} = 3x^2\dfrac{dx}{dt} ...(1).
Given \dfrac{dy}{dt} = 8\dfrac{dx}{dt}, (1) gives 48 = 3x^2 \Rightarrow x = \pm 4.
At x = 4: y = 11. At x = -4: y = -\dfrac{31}{3}.
V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}. At r=1,\ \dfrac{dr}{dt}=\tfrac{1}{2}: 4\pi(1)\left(\dfrac{1}{2}\right) = 2\pi.
r = \dfrac{3}{4}(2x+1) \Rightarrow V = \dfrac{4}{3}\pi r^3 = \dfrac{9}{16}\pi(2x+1)^3. \dfrac{dV}{dx} = \dfrac{27}{8}\pi(2x+1)^2.
Given h = \dfrac{r}{6} \Rightarrow r = 6h ...(1).
Substituting (1): V = \dfrac{1}{3}\pi r^2 h = 12\pi h^3 \Rightarrow \dfrac{dV}{dt} = 36\pi h^2\dfrac{dh}{dt}.
At h=4: 12 = 36\pi(16)\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{1}{48\pi}.
\dfrac{dC}{dx} = 0.021x^2 - 0.006x + 15. At x=17: 6.069 - 0.102 + 15 = 20.967.
\dfrac{dR}{dx} = 26x + 26. At x=7: 182 + 26 = 208.
\dfrac{dA}{dr} = 2\pi r. At r=6: 12\pi.
\dfrac{dR}{dx} = 6x + 36. At x=15: 90 + 36 = 126.
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