Question

Find the slope of the curve and the equation of tangent line of the parametric equation to the given point.

5. x=√t , y = 2t + 4 , t=1

Answer 1

Solution;

To find the equation of the tangent line, we need the slope $m=\frac{dy}{dx}$ and the point of tangency $(x_o,y_o)$

Then the equation is the usual;

$(y-y_o)=m(x-x_o)$

$x=\sqrt t,y=2t+4,t=1$

So we compute;

$\frac{dx}{dt}=\frac{1}{\sqrt t}$

$\frac{dy}{dt}=2$

Therefore;

$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=2.\sqrt t=2\sqrt t$

Now put t=1;

$m=\frac{dy}{dx}=2\sqrt 1=2$

Also ,at t=1, the original equations give;

$x_o=\sqrt t=\sqrt 1=1$

$y_o=2t+4=2(1)+4=6$

Now we put in the info for the tangent line:

$y-y_o=m(x-x_o)$

$y-6=2(x-1)$

$y=2x+4$