Question

question of differentiation

Answer 1

Find the first derivative and find the values of t for which it is zero. Then calculate the second derivative and find its value for the values of t calculates using first derivative. So, you will get the answer as ±1.

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Answer 2

Given :

$x = \dfrac{ {t}^{3} }{3} - \dfrac{5}{2} {t}^{2} + 6t + 1$

$\dfrac{dx}{dt} = 0$

To find :

$\dfrac{ {d}^{2}x }{d {t}^{2} }$

Solution :

$\implies \: x = \dfrac{ {t}^{3} }{3} - \dfrac{5}{2} {t}^{2} + 6t + 1$

Now, differentiating both sides :-

$\implies \: \dfrac{dx}{dt} = \dfrac{d(\dfrac{ {t}^{3} }{3} - \dfrac{5}{2} {t}^{2} + 6t + 1)}{dt}$

$\implies \: \dfrac{dx}{dt} = \dfrac{d(\dfrac{ {t}^{3} }{3})}{dt} - \dfrac{d( \dfrac{5}{2} {t}^{2})}{dt} + \dfrac{d(6t)}{dt} + \dfrac{d(1)}{dt}$

$\implies \: \dfrac{dx}{dt} = \dfrac{3 {t}^{2} }{3} - \dfrac{(5 \times 2)t}{2} + 6 + 0$

$\implies \: \dfrac{dx}{dt} = \dfrac{ {t}^{2} }{1} - \dfrac{5t}{1} + 6$

$\implies \: \dfrac{dx}{dt} = {t}^{2} - {5t} + 6$

now it is given that :-

$\dfrac{dx}{dt} = 0$

therefore now :-

$\implies \: 0 = {t}^{2} - {5t} + 6$

$\implies \: 0 = {t}^{2} - 2t - 3t + 6$

$\implies \: 0 = t ( {t} - 2) - 3(t - 2)$

taking out (t-2) as common :-

$\implies \: 0 = (t - 3)( {t} - 2)$

therefore, t = 3 and 2

Now :-

$\implies \: \dfrac{ {d}^{2}x }{d {t}^{2} } = \dfrac{d({t}^{2} - {5t} + 6)}{dt}$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = \dfrac{d( {t}^{2} )}{dt} - \dfrac{d(5t)}{dt} + \dfrac{d(6)}{dt}$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = 2t - 5 + 0$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = 2t - 5$

Now put t = 3 and 2 :-

Case 1 :- when t = 2

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = (2 \times 2)- 5$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = 4- 5$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = - 1$

Case 2 :- when t = 3

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = (2 \times 3)- 5$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = 6- 5$

$\implies \dfrac{ {d}^{2}x }{d {t}^{2} } = 1$

Answer :

$option \: (1) \: \pm1 \: is \: correct$