Answer:
Velocity of the first boat $\hat V_1= 10\hat i$
Velocity of the second boat $\hat V_2 = 16cos(\frac{\pi}{3})\hat i + 16sin(\frac{\pi}{3})\hat j$
After a time "t"
Displacement by boat 1 $\hat D_1 = 10t\hat i$
Displacement by boat 2 $\hat D_2 = 16tcos(\frac{\pi}{3})\hat i + 16tsin(\frac{\pi}{3})\hat j$
Using parallelogram law to find the Displacement between Boat 1 and Boat 2
$\hat D = \hat D_1 - \hat D_2$
$\hat D = (10-16cos(\frac{\pi}{3}))t\hat i - 16tsin(\frac{\pi}{3})\hat j$
Magnitude of the displacement (r) =$\sqrt{\left({\left(10t-16tcos(\frac{\pi}{3})\right)}^2 + {\left(16tsin(\frac{\pi}{3})\right)}^2\right)}$
If you simplify, then you get $r = \sqrt{\left(100t^2 + 256t^{2}{(cos(\frac{\pi}{3}))}^2 - 2.10.16t^{2}cos(\frac{\pi}{3}) + 256t^{2}{(-sin(\frac{\pi}{3}))}^2\right)}$
$ r = \sqrt{\left(100t^2 + 256t^{2} - 2.10.16t^{2}cos(\frac{\pi}{3}) \right)}$
This is where you are getting the forumula that you mentioned in the book.
$ r = \sqrt{\left(356t^2 -2.10.16.t^{2}\frac{1}{2}\right)}$
$ r = \sqrt{\left(356t^2 - 160t^{2}\right)}$
$ r = \sqrt{196t^2} = 14t$
After two hours into the jounrney, the distance would be $=14\times2 = 28$
The rate at which it is increasing would be
$dr/dt = 14$ miles per hour.
you get the "Rate at which the distance is increasing when boats are 2 hours into the journey" = $14$ miles.
This is the "Physics" way of solving the problem.
Thanks
Satish
Set of the first planes distance from the point as a function of time, do the same with the second plane. Then use the pythagorean thereom to get the distance inbetween them on terms of t. Finally differentiate that distance to get the rate of change of the distance between them.
And then it makes since that that distance is decreasing linearly because it is dependent on two linearly changing rates.
Although implicit differentiation may be quicker I would say that x=225-450t and y=300-600t. Therefore the distance between them is given by w=sqrt(x^2+y^2) which you can easily differentiate without having the derivative reference itself. Then for b, how close they come, it's an optimization problem. Find when dw/dt=0 then determine if it's a minimum (it may turn out there's no minimum and the closest they come is 0 ft apart when the graph crosses the x axis, if it does).
Best Answer
Solutions
To begin, note down the "givens": $\frac{da}{dt} = 40$, $\frac{db}{dt} = 60$. At 12pm, cars A and B are at (0,0) and (0, -90) respectively. Because A is traveling along the x-axis and B along the y-axis for the purposes of this solution, position $a$ represent's car A's x value, and $b$ is car B's y value.
Next, evaluate the cars' positions at 1pm, keeping in mind they are traveling perpendicular to each other. Cars A and B will be at (40, 0) and (0, -30) respectively, because of the velocities provided. Simply put, car A has moved "right" along the x-axis by 40, and car B has moved "up" along the y-axis by 60, because exactly one hour passed since 12pm, hence 1 * the velocity.
The first question asks us to find the rate at which the distance is changing, or $\frac{dd}{dt}$, where $d$ is the distance between cars A and B. To do this, we use the distance formula, $d = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$, where $a_1$ and $b_1$ represent car A and B's x coords, and $a_2$ and $b_2$ represent car A and B's y coords. But because car A always travels along the x-axis, and car B along the y-axis, $a_2$ (car A's y value) is always zero and $b_1$ (car B's x value) is always zero. $$d = \sqrt{(a_1 - 0)^2 + (0 - b_2)^2}$$ $$d = \sqrt{(a_1)^2 + (-b_2)^2}$$ $$d^2 = (a_1)^2 + (-b_2)^2$$ $$d^2 = (a_1)^2 + (b_2)^2$$ And finally, $d^2 = a^2 + b^2$, because using the axes simplifies the problem, as explained in the first paragraph.
Implicitly derive $$2d\frac{dd}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt}$$ $$\frac{dd}{dt} = \frac{a\frac{da}{dt} + b\frac{db}{dt}}{d}$$
Plug in $40$ for $a$, $-30$ for $b$, $40$ for $\frac{da}{dt}$, $60$ for $\frac{db}{dt}$, and $50$ for $d$ (using Pythagorean Theorem, $30^2 + 40^2 = 50^2$). $$\frac{dd}{dt} = \frac{40(40) + -30(60)}{50}$$ $$\frac{dd}{dt} = \frac{1600 -1800}{50}$$ $$\frac{dd}{dt} = \frac{-200}{50}$$ $$\frac{dd}{dt} = -4$$
Answer 1 is D
Question two concerns optimization; finding the minimum value of the distance. We know the initial positions of the cars at 12pm, and their velocities. The initial point plus the distance traveled per hour for $t$ hours tells us the positions of the cars at $t$. We can find that car A at time $t$ is at $0 + 40t$ or $40t$ and car B is at $-90 + 60t$.
Using the rate of change of the distance computed above, $$\frac{dd}{dt} = \frac{a\frac{da}{dt} + b\frac{db}{dt}}{d}$$
Plug in the values and set the derivative to zero to optimize.
$$\frac{dd}{dt} = \frac{40t(40) + (-90 + 60t)60}{d} = 0$$ $$40t(40) + (-90 + 60t)60 = 0$$ $$5200t - 5400 = 0$$ $$5200t = 5400$$ $$t=27/26$$
And perform a simple check to verify the minimum. Plugging in $t = 1$ into the numerator portion, for simplicity, yields $-200$; $t = 2$ yields $5000$. The derivative is 0 and is increasing at $27/26$, so the distance is indeed minimized at that time.
Answer for 2 is B
Feel free to offer things to clarify, add, or remove, as this is my first lengthy and sophisticated solution.