Question

application of derivatives find the equation of tangent and normal to the point on it x=√t y=t-1/√t at t =4​

Answer 1

Answer:

$\cfrac{dy}{dx} = \cfrac{t+1}{t}$

Step-by-step explanation:

$x=\sqrt{t}$

Differentiating w.r.t. t,

$\cfrac{dx}{dt} = \cfrac{1}{2\sqrt{t}}$

Now, $y = \cfrac{t-1}{\sqrt{t}}$

Differentiating w.r.t. t,

$\cfrac{dy}{dt} = \cfrac{\sqrt{t} \cfrac{d}{dt}(t-1) - (t-1)\cfrac{d}{dt}\sqrt{t}}{(\sqrt{t})^2}$

$= \cfrac{\sqrt{t}(1) - \cfrac{(t-1)}{2\sqrt{t}}}{t}$

$= \cfrac{2t-(t-1)}{2\sqrt{t} \cdot t}$

$= \cfrac{2t-t+1}{2 t^{3/2}}$

$= \cfrac{t+1}{2t^{3/2}}$

Now, slope of tangent is $\cfrac{dy}{dx}$

So,

$\cfrac{dy}{dx} = \cfrac{t+1}{2 t^{3/2}} \times 2\sqrt{t}\\= \cfrac{t+1}{t^{3/2}} \times t^{1/2}$

$= (t+1) \times t^{1/2 - 3/2}$

$= (t+1) \times t^{-1/2}$

$\therefore \cfrac{dy}{dx} = \cfrac{t+1}{t}$

Now, equation of tangent is given by

$y- y_1 = m(x - x_1)$

Substituting the respective values of x1, y1 and m (slope of tangent), we get

$y-x \left(\cfrac{t+1}{t}\right) = \cfrac{-2}{\sqrt{t}}$

Now, slope of normal

$\cfrac{dx}{dy} = -\cfrac{t}{t+1}$

So, equation of normal is

$y - \cfrac{t}{t+1}x = \cfrac{2t^2 - 1}{(t+1)\sqrt{t}}$