Question

find dx/dy of y =at² & x=at​

Answer 1

Answer:

$\rm \dfrac{1}{2t}$

Step-by-step explanation:

Given :

y = at² & x = at​

To find :

$\mathrm{\dfrac{dx}{dy}}$

Solution :

Given, y = at² and x = at​

Differentiating x with respect to t, we get,

$\implies \mathrm{\dfrac{dx}{dt} = \dfrac{d}{dt}(at)}$

We know that,

$\boxed{\mathrm{\dfrac{d}{dx}(x) = 1}}$

So,

$\implies \mathrm{\dfrac{dx}{dt} = a}$ ___ [1]

Now, Differentiating y with respect to t, we get,

$\implies \mathrm{\dfrac{dy}{dt} = \dfrac{d}{dt}(at^2)}$

We know that,

$\boxed{\mathrm{\dfrac{d}{dx}(x^n) = nx^{n-1}}}$

here, n = 2,

So, n - 1 = 2 - 1 = 1

Thus,

$\implies \mathrm{\dfrac{dy}{dt} = 2at}$ ___[2]

Now, dividing [1] by [2], we get,

$\implies \mathrm{\dfrac{\bigg(\dfrac{dx}{dt}\bigg)}{\bigg(\dfrac{dy}{dt}\bigg)} = \dfrac{a}{2at}}$

$\implies \mathrm{\dfrac{\bigg(\dfrac{dx}{\not{dt}}\bigg)}{\bigg(\dfrac{dy}{\not{dt}}\bigg)} = \dfrac{\not{a}}{2\not{a}t}}$

$\implies \mathrm{\dfrac{dx}{dy} = \dfrac{1}{2t}}$

$\therefore \underline{\boxed{\mathbf{\dfrac{dx}{dy} = \dfrac{1}{2t}}}}$

❖ Extra information :

⟡ Some basic differentiations :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf x^n & \sf nx^{n-1} \\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x} \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx \end{array}} \\ \end{gathered}$

Hope it helps!!