find an equation of the tangent to the curve at the point corresponding to the given value of the parameter…

find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t^4 + 5 y = t^3 + t t = 1 y(x) =

Answer

When $t = 1$, $x=1^{4}+5=6$ and $y = 1^{3}+1=2$. So the point is $(6,2)$.

First, find $\frac{dx}{dt}=4t^{3}$ and $\frac{dy}{dt}=3t^{2}+1$. Then $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^{2}+1}{4t^{3}}$.

Substitute $t = 1$ into $\frac{dy}{dx}$, we get $\frac{dy}{dx}\big|_{t = 1}=\frac{3\times1^{2}+1}{4\times1^{3}} = 1$.

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(6,2)$ and $m = 1$. So $y-2=1\times(x - 6)$.

$y-2=x - 6$, which simplifies to $y=x - 4$.

$y=x - 4$