d72b08…: “Versus the dark owl”

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Suppose that a bug walks along the parabola $y=x^2+3$ and does so in a manner in which $\frac {dx}{dt} = 3$ . If the bug starts walking at $x=1$ at time $t=0$ , and we want $x(t)$ to measure the bug’s $x$ -coordinate at time $t$ , then we have

$x(t) = \answer {3t+1}.$

Since $\frac {dx}{dt} = 3$ , we can integrate to obtain

$x(t) = \answer {3t+C}$

(use $C$ as the constant of integration).

Now, since $x(0) = 1$ , $C=\answer {1}$ .

Suppose that we now want to use a vector-valued function to give the position of the bug at time $t$ . Then, $y(t) = \answer {9t^2+6t+4}$ and the position vector is

$\vec {p}(t) = \vector {x(t),y(t)} = \vector {\answer {3t+1},\answer {9t^2+6t+4}}, t \geq 0.$

Since $y=x^2+3$ and we have that $x(t) = 3t+1$ , we can find $y(t)$ by substituting the expression for $x(t)$ into the equation for the parabola.

Try to imagine the bug’s motion subject to these conditions (or use the instructions in the parametric equations homework to design a Desmos sheet). As $t>0$ , we should have that $\frac {dy}{dt}$ is increasingdecreasingconstant .

Now, find the time $t$ when $\frac {dy}{dt} = 60$ . How far is the bug away from the origin at this time?

$\frac {dy}{dt} = 60$ when $t=\answer {3}$ , and the bug is $\answer {\sqrt {10709} }$ units from the origin at this time.

We can compute $\frac {dy}{dt}$ from $y(t) = 9t^2+6t+4$ to find $\frac {dy}{dt} = \answer {18t+6}$ . Setting this equal to $60$ , we find that $t=\answer {3}$ .

To find the distance from the origin the bug is at this time, note that we can find the position of the bug from evaluating $\vec {p}(t)$ at $t=3$ , and the distance from the origin will be exactly $|\vec {p}(3)|$ .