Question

Derive the equation: $u = {x^2} + {y^2}$ when $x = s + 3t$ and $y = 2s - t$ . What will be the value of $\dfrac{{{d^2}u}}{{d{s^2}}}$
(a) $10$
(b) $11$
(c) $12$
(d) None of the above

Answer 1

Hint : We will use the most eccentric concept of derivations. Assuming the given satisfied equation $f(u) = {x^2} + {y^2}$ for the ease of the problem. Considering the given parameters i.e. $s{\text{ and }}t$ derivative these equations twice, with respect to $s$ given in the problem. As a result, substituting the values in the respective calculation/s the desired value is obtained.

Complete step-by-step answer :
Since, we have the given equation
$u = {x^2} + {y^2}$
Let us assume, $f\left( u \right)$ be the function of ${x^2} + {y^2}$
$\therefore f\left( u \right) = {x^2} + {y^2}$
First of all, derivating the given parameter $x = s + 3t$ with respect to ‘$s$’ variable solving the equation,
$\Rightarrow \dfrac{{dx}}{{ds}} = 1 + 0 = 1$
Similarly, derivating the second parameter $y = 2s - t$ with respect to ‘$s$’ variable solving the equation,
$\Rightarrow \dfrac{{dy}}{{ds}} = 2 + 0 = 2$
Since, derivation of any constant is always ‘zero’ that is $\dfrac{d}{{dx}}\left( t \right) = 0$ (here, constant is ‘$t$’ respectively)
Again, derive the above equations with respect to ‘$s$’, we get
$\Rightarrow \dfrac{{{d^2}x}}{{d{s^2}}} = 0$
And,
$\Rightarrow \dfrac{{{d^2}y}}{{d{s^2}}} = 0$
But, we have given that$f(u)$satisfies these parameters, we get
$f\left( u \right) = {x^2} + {y^2}$
Derive the equation with respect to ‘$s$’, we get
$\Rightarrow f'\left( u \right) = \left( {2x} \right)\dfrac{{dx}}{{ds}} + \left( {2y} \right)\dfrac{{dy}}{{ds}}$
Where,
$f'\left( u \right) = \dfrac{{du}}{{ds}}$
Again, derive the above equations with respect to ‘$s$’, we get
$\Rightarrow f''\left( u \right) = 2\left( {x\dfrac{{{d^2}x}}{{d{s^2}}} + \dfrac{{dx}}{{ds}}} \right) + 2\left( {y\dfrac{{{d^2}y}}{{d{s^2}}} + \dfrac{{dy}}{{ds}}} \right)$
Where,
$f''\left( u \right) = \dfrac{{{d^2}u}}{{d{s^2}}}$
Now, since substituting the values of
$\dfrac{{dx}}{{ds}} = 1$,
$\dfrac{{dy}}{{ds}} = 2$,
$\dfrac{{{d^2}x}}{{d{s^2}}} = 0$, and
$\dfrac{{{d^2}y}}{{d{s^2}}} = 0$, we get
$\Rightarrow f''\left( u \right) = 2\left( {0 + 1} \right) + 2\left( {0 + 2} \right) = 2 + 2\left( 4 \right)$
$\Rightarrow f''\left( u \right) = 10$
So, the correct answer is “Option a”.

Note : One must remember the concept of derivation, how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as $\dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$ is the multiplication rule of derivation used here and $\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}},...$ can be noted as and so on! Derivation of any constant number is always zero. Deriving the equation with the same term is always one $\dfrac{d}{{dx}}(x) = 1$ . An algebraic identity plays a significant role in solving this problem.