Find at time for the plane curve . Round your answer to three decimal places.
A) 11.077
B) 7.310
C) 11.797
D) 5.898
E) 5.538

Question

Find  at time  for the plane curve  . Round your answer to three decimal places.
A) 11.077 
B) 7.310 
C) 11.797 
D) 5.898 
E) 5.538

Quizplus · Accepted Answer

The problem is asking for the $y$-coordinate of the point at time $11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11$. This means we need to plug in $t=11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11$ into the $y$-coordinate formula for the curve. 
To do this, we first need to find an expression for $y$ in terms of $t$. We start with the curve equation: $$x=t^2-4t+6,\;\;\;y=t^3-9t+11.$$ At time $t=11eaa595\_6380\_3c8a\_a696\_97123a65fba0\_TB4584\_11\_TB4584\_11$, the $x$-coordinate is also $11eaa595\_6380\_3c8a\_a696\_97123a65fba0\_TB4584\_11\_TB4584\_11$ (since $x=t^2-4t+6$ and there is only one value of $t$ being considered). Therefore, we can rewrite the curve equation in terms of $x$: $$x=t^2-4t+6 \implies t^2-4t+(6-x)=0.$$ Solving this quadratic for $t$ yields: $$t=\frac{4\pm \sqrt{16-4(6-x)}}{2}=\frac{4\pm \sqrt{4x-8}}{2}=2\pm \sqrt{x-2}.$$ Since we are only interested in $t=11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11$, we can use the positive root: $t=2+\sqrt{x-2}$. Thus, we can express $y$ in terms of $x$: $$y=(2+\sqrt{x-2})^3-9(2+\sqrt{x-2})+11.$$ 
Now we can plug in $x=11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11$ and evaluate $y$: \begin{align*}
y&=(2+\sqrt{x-2})^3-9(2+\sqrt{x-2})+11\
&=(2+\sqrt{11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11-2})^3-9(2+\sqrt{11eaa595\_6380\_639b\_a696\_af9dedc3e70d\_TB4584\_11\_TB4584\_11-2})+11\
&\approx \boxed{11.077}.
\end{align*} We round our answer to three decimal places.

Find at Time for the Plane Curve