Question

section 2.3 score: 8/9 answered: 8/9 question 9 youre flying from joint base lewis - mcchord (jblm) to an undisclosed location 170 km south and 139 km east. mt. rainier is located approximately 56 km east and 40 km south of jblm. if you are flying at a constant speed of 800 km/hr, how long after you depart jblm will you be the closest to mt. rainier? minutes question help: post to forum

Explanation:

Step1: Define position - vector of the flight path

Let the position of JBLM be the origin \((0,0)\). The direction - vector of the flight path from JBLM to the undisclosed location is \(\vec{v}=(139, - 170)\) (east is the positive x - direction and south is the negative y - direction). The position of Mt. Rainier is \(\vec{r}_{0}=(56,-40)\). The position of the plane at time \(t\) (assuming it starts at \(t = 0\)) is \(\vec{r}(t)=t\vec{v}=(139t,-170t)\).

Step2: Calculate the distance - vector

The distance - vector between the plane and Mt. Rainier is \(\vec{d}(t)=(139t - 56,-170t + 40)\).

Step3: Square the distance function

The square of the distance \(D(t)\) between the plane and Mt. Rainier is \(D(t)=(139t - 56)^{2}+(-170t + 40)^{2}\). Expand it:
\[

$$\begin{align*} D(t)&=(139t - 56)^{2}+(-170t + 40)^{2}\\ &=139^{2}t^{2}-2\times139\times56t + 56^{2}+170^{2}t^{2}-2\times170\times40t + 40^{2}\\ &=(139^{2}+170^{2})t^{2}-(2\times139\times56 + 2\times170\times40)t+(56^{2}+40^{2}) \end{align*}$$

\]
where \(139^{2}=19321\), \(170^{2}=28900\), \(2\times139\times56 = 2\times7784=15568\), \(2\times170\times40 = 13600\), \(56^{2}=3136\), \(40^{2}=1600\). So \(D(t)=(19321 + 28900)t^{2}-(15568+13600)t+(3136 + 1600)=48221t^{2}-29168t + 4736\).

Step4: Differentiate the distance - squared function

Differentiate \(D(t)\) with respect to \(t\): \(D^\prime(t)=2\times48221t-29168 = 96442t-29168\).

Step5: Set the derivative equal to zero

Set \(D^\prime(t)=0\) to find the critical points:
\[

$$\begin{align*} 96442t-29168&=0\\ t&=\frac{29168}{96442}\text{ hours} \end{align*}$$

\]

Step6: Convert time to minutes

Since \(1\) hour \( = 60\) minutes, \(t=\frac{29168}{96442}\times60\) minutes.
\[

$$\begin{align*} t&=\frac{29168\times60}{96442}\\ &=\frac{1750080}{96442}\\ &\approx18.15\text{ minutes} \end{align*}$$

\]

Answer:

\(18.15\)