If x = at2, y = 2 at, then \[\frac{d^2 y}{d x^2} =\]

(d) \[- \frac{1}{2 \ a t^3}\] \[x = a t^2 \text { and y } = 2\text { at }\]\[\text { Differentiating w . r . t . t, we get }\]\[\frac{d x}{d t} = 2\text { at and } \frac{d y}{d t} = 2a\]\[ \therefore \frac{d y}{d x} = \frac{2a}{2at} = \frac{1}{t}\]\[\text{ Differentiating again w . r . t . t, we get }\]\[\frac{d^2 y}{d x^2} = \frac{- 1}{t^2}\frac{d t}{d x} = \frac{- 1}{2 \ a t^3}\]

If X = At2, Y = 2 At, Then D 2 Y D X 2 = (A) − 1 T 2 - Mathematics

Question

If x = at², y = 2 at, then $\frac{d^{2} y}{d x^{2}} =$

Options

$- \frac{1}{t^{2}}$
$\frac{1}{2 a t^{3}}$
$- \frac{1}{t^{3}}$
$- \frac{1}{2 a t^{3}}$

MCQ

Solution

(d) $- \frac{1}{2 a t^{3}}$

$x = a t^{2} and y = 2 at$
$Differentiating w . r . t . t, we get$
$\frac{d x}{d t} = 2 at and \frac{d y}{d t} = 2 a$
$∴ \frac{d y}{d x} = \frac{2 a}{2 a t} = \frac{1}{t}$
$Differentiating again w . r . t . t, we get$
$\frac{d^{2} y}{d x^{2}} = \frac{- 1}{t^{2}} \frac{d t}{d x} = \frac{- 1}{2 a t^{3}}$

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Simple Problems on Applications of Derivatives

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Chapter 12: Higher Order Derivatives - Exercise 12.3 [Page 22]

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Chapter 12 Higher Order Derivatives
Exercise 12.3 | Q 2 | Page 22

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If X = At2, Y = 2 At, Then D 2 Y D X 2 = (A) − 1 T 2 - Mathematics

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