Answer
(a) The equation of the tangent line is $~~y = 2x+1$
(b) The equation of the tangent line is $~~y = 2x+1$
Work Step by Step
(a) The equation of the tangent line is $\frac{dy}{dx}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2t}{1/t} = 2t^2$
When $x=1$:
$x = 1+ln~t = 1$
$ln~t = 0$
$t=1$
The slope of the tangent line is $2(1)^2 = 2$
We can find the equation of the tangent line:
$y-3 = 2(x-1)$
$y = 2x+1$
(b) $x = 1+ln~t$
$ln~t = x-1$
$t = e^{x-1}$
We can replace $t$ in the expression for $y$:
$y = t^2+2$
$y = (e^{x-1})^2+2$
$y = e^{2x-2}+2$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = 2e^{2x-2}$
When $x=1$:
$x = 1+ln~t = 1$
$ln~t = 0$
$t=1$
The slope of the tangent line is $2e^{2(1)-2} = 2$
We can find the equation of the tangent line:
$y-3 = 2(x-1)$
$y = 2x+1$
