find an equation for the tangent line to the curve parametrized by $x = t^{3}+1$ and $y = t^{2}+t$ at the…

Explanation:

Step1: Find the coordinates of the point

When $t = 2$, $x=t^{3}+1=2^{3}+1=8 + 1=9$, $y=t^{2}+t=2^{2}+2=4 + 2=6$. So the point on the curve is $(9,6)$.

Step2: Find $\frac{dy}{dx}$

First, find $\frac{dx}{dt}=3t^{2}$ and $\frac{dy}{dt}=2t + 1$. Then by the chain - rule $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t + 1}{3t^{2}}$.

Step3: Find the slope of the tangent line

Substitute $t = 2$ into $\frac{dy}{dx}$. $\frac{dy}{dx}\big|_{t = 2}=\frac{2\times2+1}{3\times2^{2}}=\frac{4 + 1}{12}=\frac{5}{12}$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(9,6)$ and $m=\frac{5}{12}$. $y-6=\frac{5}{12}(x - 9)$ $y-6=\frac{5}{12}x-\frac{15}{4}$ $y=\frac{5}{12}x-\frac{15}{4}+6$ $y=\frac{5}{12}x+\frac{9}{4}$

Answer:

$y=\frac{5}{12}x+\frac{9}{4}$