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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 1–10 (part 1 of 7)

  1. Exercise 1

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

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    Let \(\displaystyle r\) be the radius and \(\displaystyle k>0\) the constant of proportionality.\[V = \tfrac43\pi r^3, \qquad S = 4\pi r^2 \]\[-\frac{dV}{dt} = kS = 4\pi k r^2 \quad \text{(given)} \]\[\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt} \]\[4\pi r^2\,\frac{dr}{dt} = -4\pi k r^2 \]\[\frac{dr}{dt} = -k \]Answer: \(\displaystyle \dfrac{dr}{dt} = -k\), a constant, so the radius decreases at a constant rate.

  2. Exercise 2

    If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle r\) be the radius, \(\displaystyle A\) the area, \(\displaystyle P\) the perimeter, and \(\displaystyle \dfrac{dA}{dt} = c\) a constant.\[A = \pi r^2 \]\[\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt} = c \]\[\frac{dr}{dt} = \frac{c}{2\pi r} \]\[P = 2\pi r \;\Rightarrow\; \frac{dP}{dt} = 2\pi\,\frac{dr}{dt} \]\[\frac{dP}{dt} = 2\pi\cdot\frac{c}{2\pi r} = \frac{c}{r} \]Answer: \(\displaystyle \dfrac{dP}{dt} = \dfrac{c}{r}\), so the rate at which the perimeter changes varies inversely as the radius.

  3. Exercise 3

    A kite is moving horizontally at a height of 151.5\displaystyle 151.5 meters. If the speed of kite is 10 m/s\displaystyle 10 \mathrm{~m} / \mathrm{s}, how fast is the string being let out; when the kite is 250\displaystyle 250 m away from the boy who is flying the kite? The height of boy is 1.5\displaystyle 1.5 m.

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    NCERT’s answer
    $\displaystyle 8$ m/s
    Take the string to start at the top of the boy, point \(\displaystyle B\); \(\displaystyle K\) is the kite and \(\displaystyle P\) the point of \(\displaystyle B\)'s level directly below it.\[KP = 151.5 - 1.5 = 150 \text{ m} \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q3Let \(\displaystyle BP = x\), \(\displaystyle BK = l\).\[l^2 = x^2 + 150^2 \quad \text{(Pythagoras)} \]\[2l\,\frac{dl}{dt} = 2x\,\frac{dx}{dt} \]For \(\displaystyle l = 250\):\[x = \sqrt{250^2 - 150^2} = 200 \]\[\frac{dl}{dt} = \frac{x}{l}\cdot\frac{dx}{dt} = \frac{200}{250}\times 10 = 8 \]Answer: The string is let out at \(\displaystyle 8\) m/s.

  4. Exercise 4

    Two men A and B start with velocities v\displaystyle v at the same time from the junction of two roads inclined at 45\displaystyle 45° to each other. If they travel by different roads, find the rate at which they are being seperated..

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    NCERT’s answer
    \(\displaystyle (\sqrt{2-\sqrt{2}}) v\) unit/sec.
    Let \(\displaystyle O\) be the junction. After time \(\displaystyle t\) the men are at \(\displaystyle A\) and \(\displaystyle B\).\[OA = OB = vt, \qquad \angle AOB = 45^\circ \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q4\[AB^2 = OA^2 + OB^2 - 2\,OA\cdot OB\cos 45^\circ \quad \text{(cosine rule)} \]\[AB^2 = 2v^2t^2 - \sqrt2\,v^2t^2 = (2-\sqrt2)\,v^2t^2 \]\[AB = vt\sqrt{2-\sqrt2} \]\[\frac{d(AB)}{dt} = v\sqrt{2-\sqrt2} \]Answer: They separate at the rate \(\displaystyle v\sqrt{2-\sqrt2}\) (about \(\displaystyle 0.77v\)).

  5. Exercise 5

    Find an angle θ,0<θ<π2\displaystyle \theta, 0<\theta<\frac{\pi}{2}, which increases twice as fast as its sine.

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    NCERT’s answer
    \(\displaystyle \theta=\frac{\pi}{3}\)
    \[\frac{d\theta}{dt} = 2\,\frac{d}{dt}(\sin\theta) \quad \text{(given)} \]\[\frac{d\theta}{dt} = 2\cos\theta\,\frac{d\theta}{dt} \]Since \(\displaystyle \theta\) is increasing, \(\displaystyle \dfrac{d\theta}{dt}\neq 0\):\[1 = 2\cos\theta \;\Rightarrow\; \cos\theta = \frac12 \]\[\theta = \frac{\pi}{3} \quad \left(0<\theta<\frac{\pi}{2}\right) \]Answer: \(\displaystyle \theta = \dfrac{\pi}{3}\)

  6. Exercise 6

    Find the approximate value of (1.999)5\displaystyle (1.999)^5.

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    NCERT’s answer
    31.$\displaystyle 92$
    Take \(\displaystyle f(x) = x^5\) with \(\displaystyle x = 2\), \(\displaystyle \Delta x = 1.999 - 2 = -0.001\).\[f'(x) = 5x^4 = 5\cdot 2^4 = 80 \]\[\Delta y \approx f'(x)\,\Delta x = 80\times(-0.001) = -0.08 \]\[(1.999)^5 = f(2) + \Delta y \approx 32 - 0.08 = 31.92 \]Answer: \(\displaystyle (1.999)^5 \approx 31.92\)

  7. Exercise 7

    Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3\displaystyle 3 cm and 3.0005\displaystyle 3.0005 cm, respectively.

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    NCERT’s answer
    \(\displaystyle 0.018 \pi \mathrm{cm}^3\)
    \[V = \tfrac43\pi r^3 \;\Rightarrow\; dV = 4\pi r^2\,dr \]\[r = 3, \qquad dr = 3.0005 - 3 = 0.0005 \]Volume of metal \(\displaystyle = V(3.0005) - V(3) \approx dV\):\[dV = 4\pi(3)^2(0.0005) = 0.018\pi \]\[0.018\pi \approx 0.0565 \]Answer: Volume of metal \(\displaystyle \approx 0.018\pi \approx 0.057\ \text{cm}^3\).

  8. Exercise 8

    A man, 2m tall, walks at the rate of 123 m/s\displaystyle 1 \frac{2}{3} \mathrm{~m} / \mathrm{s} towards a street light which is 513 m\displaystyle 5 \frac{1}{3} \mathrm{~m} above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is 313 m\displaystyle 3 \frac{1}{3} \mathrm{~m} from the base of the light?

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    NCERT’s answer
    \(\displaystyle 2 \frac{2}{3} \mathrm{~m} / \mathrm{s}\) towards light, \(\displaystyle -1 \mathrm{~m} / \mathrm{s}\)
    Let \(\displaystyle OL\) be the lamp-post, \(\displaystyle MN\) the man, \(\displaystyle T\) the tip of the shadow. Put \(\displaystyle OM = x\), \(\displaystyle MT = s\), \(\displaystyle OT = y = x+s\).\[OL = \tfrac{16}{3}, \qquad MN = 2, \qquad \frac{dx}{dt} = -\frac53 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q8\[\triangle TMN \sim \triangle TOL \;\Rightarrow\; \frac{s}{x+s} = \frac{MN}{OL} = \frac{2}{16/3} = \frac38 \]\[8s = 3x + 3s \;\Rightarrow\; s = \frac35\,x, \qquad y = \frac85\,x \]\[\frac{dy}{dt} = \frac85\cdot\left(-\frac53\right) = -\frac83 \]\[\frac{ds}{dt} = \frac35\cdot\left(-\frac53\right) = -1 \]Answer: The tip moves towards the lamp at \(\displaystyle \dfrac83\) m/s; the shadow shortens at \(\displaystyle 1\) m/s, at \(\displaystyle x = \dfrac{10}{3}\) m as at any other position.

  9. Exercise 9

    A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t\displaystyle t seconds after the pool has been plugged off to drain and L=200(10−t)2\displaystyle \mathrm{L}=200(10-t)^2. How fast is the water running out at the end of 5\displaystyle 5 seconds? What is the average rate at which the water flows out during the first 5\displaystyle 5 seconds?

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    NCERT’s answer
    $\displaystyle 2000$ litres/s, $\displaystyle 3000$ litre/s
    \[L = 200(10-t)^2 \]\[\frac{dL}{dt} = -400(10-t) \]\[\left.\frac{dL}{dt}\right|_{t=5} = -400(10-5) = -2000 \]The negative sign means the pool is emptying.\[L(0) = 200(10)^2 = 20000, \qquad L(5) = 200(5)^2 = 5000 \]\[\text{average rate} = \frac{L(5)-L(0)}{5-0} = \frac{5000-20000}{5} = -3000 \]Answer: at \(\displaystyle t=5\) s the water runs out at \(\displaystyle 2000\) litres/s; during the first $\displaystyle 5$ s it averages \(\displaystyle 3000\) litres/s.

  10. Exercise 10

    The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let the side be \(\displaystyle x\) and the constant rate of volume increase be \(\displaystyle k\).\[V = x^3 \;\Rightarrow\; \frac{dV}{dt} = 3x^2\frac{dx}{dt} = k \]\[\frac{dx}{dt} = \frac{k}{3x^2} \]\[S = 6x^2 \;\Rightarrow\; \frac{dS}{dt} = 12x\,\frac{dx}{dt} \]\[\frac{dS}{dt} = 12x\cdot\frac{k}{3x^2} = \frac{4k}{x} \]Answer: \(\displaystyle \dfrac{dS}{dt} = \dfrac{4k}{x}\) with \(\displaystyle 4k\) constant, so the increase in surface area varies inversely as the side.