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.
Let $v(t)$ be the height (vertical distance} of the balloon at time $t$, where $t$ is measured in seconds. Then $$v(t) = 65 + (1\text{ ft./sec}) \;\cdot t \;\text{sec} = 65 + t \;\text{ft.}.$$
Let $h(t)$ be the horizontal distance of the bike at time $t$, where $t$ is measured in seconds. Then $$h(t) = (17 \;\text{ft/sec}) \cdot t \;\text{sec}\; = 17 t \;\text{ft.}$$
The distance between the balloon and the bicycle at time $t$ seconds is the hypotenuse formed of the triangle formed by the vertical distance and horizontal distance, as legs:
$$d(t) = \sqrt{(h(t))^2 + (v(t))^2} = \sqrt{(65 + t)^2 + (17t)^2}$$
Find $d'(t)$ and once you've done that, evaluate at $t = 3\;\text{sec.}$
Best Answer
u is the hypotenuse so
$u=\sqrt {1800^2+2400^2}$
$u^2=100^2*(18^2+24^2)$
$u^2=100^2*(900)$