Disk problem: This one is more mechanical, so let's do it first. Let $r=r(t)$ be the radius of one of its faces. Note that $r$ is changing. We are told how fast $r$ is changing. In symbols, what we are told amounts to
$$\frac{dr}{dt}=0.02.$$
We are asked how fast the area of a face (side, like a coin) is changing. Let $A=A(t)$ be the area of the face. We want to find out about $\dfrac{dA}{dt}$.
We are given the rate of change of something, and want to find the rate of change of something else. We therefore need a link between the two quantities. In our case, the link is through the familiar formula
$$A=\pi r^2.$$
Differentiate both sides with respect to $t$. To differentiate $r^2$, use the Chain Rule. We get
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}.$$
At the instant when $r=8.1$, we know everything on the right-hand side. At that instant, area is changing at the rate of $(2\pi)(8.1)(0.02)$ (square inches per second).
Plane problem: First step: Draw a diagram. Let $O$ be the position of the observer, and let $A$ be the point $1$ mile above the observer. Let $P$ be the position of the plane. The position $P$ is changing. Note that $\triangle OAP$ is right-angled.
Let $x=x(t)$ be the distance $OP$. This distance is changing. We know at what rate $x$ is changing: it is the speed of the plane. Do we know that
$$\frac{dx}{dt}=400.$$
We are asked how fast the distance of the plane from the observer is changing. So we are asked how fast $OP$ is changing. Let $OP=z$. The $z=z(t)$ is a function of $t$. We want to find out about
$$\frac{dz}{dt}.$$
We need a link between $x$ and $z$. In this case, that is provided by the Pythagorean Theorem. Since $OA=1$, we have
$$z^2=1^2+x^2.$$
Differentiate immediately* with respect to $t$, using the Chain Rule. Now you should be able to finish quickly. We get
$$2\frac{dz}{dt}=2x\frac{dx}{dt},\quad\text{or equivalently} \frac{dz}{dt}=\frac{x}{z}\frac{dx}{dt}.$$
Now freeze the situation $45$ seconds after the plane passed overhead. We can compute $x$ at that instant, and, using the Pythagorean Theorem, $z$, and now we know everything.
Note Many students would write instead that $z=\sqrt{1+x^2}$, and then differentiate. That's perfectly fine, a bit more work, a bit greater chance of error.
The average rate of change of $f(t)$ on the interval $a\le t\le b$ is just ${f(b)-f(a)\over b-a}$. So here---assuming I am interpreting your function correctly with the extra set of parentheses---you get:
$${f(4)-f(1)\over 4-1}\approx -0.373338.$$
To understand this geometrically, just visualize the secant line between the points $(1,f(1))$ and $(4,f(4))$ shown in black. The number above is the slope of this line. The original function is shown in blue. (Note how the axes are scaled though.)
Best Answer
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