Answer
(a) The equation of the tangent line is $~~y = 4e~x-7e$
(b) The equation of the tangent line is $~~y = 4e~x-7e$
Work Step by Step
(a) The equation of the tangent line is $\frac{dy}{dx}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2te^{t^2}}{1/2\sqrt{t}} = 4t^{3/2}e^{t^2}$
When $x=2$:
$x = 1+\sqrt{t} = 2$
$\sqrt{t} = 1$
$t=1$
The slope of the tangent line is $4(1)^{3/2}e^{(1)^2}= 4e$
We can find the equation of the tangent line:
$y-e = 4e(x-2)$
$y = 4e~x-7e$
(b) $x = 1+\sqrt{t}$
$\sqrt{t} = x-1$
$t = (x-1)^2$
We can replace $t$ in the expression for $y$:
$y = e^{t^2}$
$y = e^{[(x-1)^2~]^2}$
$y = e^{(x-1)^4}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = 4(x-1)^3~e^{(x-1)^4}$
The slope of the tangent line is $4(2-1)^3~e^{(2-1)^4}= 4e$
We can find the equation of the tangent line:
$y-e = 4e(x-2)$
$y = 4e~x-7e$
