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.
What you calculated is the average rate of change of area between radii of 100 and 120 cm, not the instantaneous rate at 120 cm.
To solve this problem, note that $\frac d{dr}\pi r^2$ (derivative of the circle's area with respect to its radius) is $2\pi r$, and substituting $r=120$ yields the correct answer of $240\pi$.
Best Answer
First note that the area of the circular ripple is $A=\pi r^2$
$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$
From the question we have, $\dfrac{dr}{dt}=6$ inches per second
So, the rate of change of area with respect to time is $$\dfrac{dA}{dt}=2\pi r(6)=12\pi r$$
Now, to find the rates of areas with the radius you have, you need to plug in the values of $r$ in the above area rate equation.
Can you take it from here?