Question

How do you find $\dfrac{{{d^2}y}}{{d{x^2}}}$ for the curve $x = 4 + {t^2}$ and $y = {t^2} + {t^3}$?

Answer 1

Here we need to double differentiate the function $y$ with respect to $x$ but here we are given $y{\text{ and }}x$ in term of $t$ so we can differentiate $y{\text{ and }}x$ with respect to $t$ and then divide them both to get $\dfrac{{dy}}{{dx}}$ and again differentiate to get the required result.

Complete step by step solution:
Here we are given $x = 4 + {t^2}$ and $y = {t^2} + {t^3}$ but we need to find the $\dfrac{{dy}}{{dx}}$.
If we had $y$ in terms of $x$ then we would have differentiated it directly with respect to $x$ directly but here we have $y{\text{ and }}x$ in term of $t$ so we can differentiate $y{\text{ and }}x$ with respect to $t$ and then divide them both to get $\dfrac{{dy}}{{dx}}$ and again differentiate to get the required result.
So we have
$y = {t^2} + {t^3}$ $- - - (1)$
Differentiating the equation (1) with respect to $t$ we will get:
$\dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}({t^2} + {t^3}) \\\ \Rightarrow \dfrac{{dy}}{{dt}} = 2t + 3{t^2} - - - - (2) \\\$
Again we have $x = 4 + {t^2}$ and we can again differentiate with respect to $t$
$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}(4 + {t^2}) \\\ \Rightarrow \dfrac{{dx}}{{dt}} = 2t - - - - (3) \\\$
Dividing equation (2) by equation (3) we will get:
$\Rightarrow$ $\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{2t + 3{t^2}}}{{2t}} = 1 + \dfrac{{3t}}{2}$
So we get:
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = 1 + \dfrac{{3t}}{2}$
Now we need to find the value of double differentiation which is $\dfrac{{{d^2}y}}{{d{x^2}}}$
Now we can write $\dfrac{{{d^2}y}}{{d{x^2}}}$ as $\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)$
Now if we put the value of $\dfrac{{dy}}{{dx}} = 1 + \dfrac{{3t}}{2}$
We will get:
$\Rightarrow$ $\dfrac{{{d^2}y}}{{d{x^2}}}$ $= \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {1 + \dfrac{{3t}}{2}} \right)$
Now dividing numerator and denominator by $dt$, we will get:
$\Rightarrow$ $\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{\dfrac{d}{{dt}}\left( {1 + \dfrac{{3t}}{2}} \right)}}{{\dfrac{{dx}}{{dt}}}}$ $- - - - - (4)$
Now as we know the value of differentiation of $x$ with respect to $t$ from the equation (3) we can substitute the value of $\dfrac{{dx}}{{dt}} = 2t$ in equation (4) we will get:
$\Rightarrow$ $\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{\dfrac{d}{{dt}}\left( {1 + \dfrac{{3t}}{2}} \right)}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{\dfrac{d}{{dt}}\left( {1 + \dfrac{{3t}}{2}} \right)}}{{2t}}$
Now we can differentiate the above numerator easily and get:
$\Rightarrow$ $\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{\dfrac{d}{{dt}}\left( {1 + \dfrac{{3t}}{2}} \right)}}{{2t}} = \dfrac{{\dfrac{3}{2}}}{{2t}} = \dfrac{3}{{4t}}$

Hence we get the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$ $= \dfrac{3}{{4t}}$.

Note: Here in these types of problems the student can make a mistake by finding $\dfrac{{dy}}{{dx}}$ and then differentiating it twice with respect to $t$ only but we firstly need to divide numerator and denominator by $dt$ and then only we need to make further step as we have done above.