let $x$ be the horizontal component and $y$ the vertical component of the kite of length $k$, then from Pythagoras' $$k^2=x^2+y^2$$
Apply implicit differentiation with respect to time and you get $$2k\cdot\cfrac{dk}{dt}= 2x\cdot\cfrac{dx}{dt}+ 2y\cdot\cfrac{dy}{dt} $$
The kite flies only horizontally, thus there is no variation of $y$ with respect to time and $\cfrac{dy}{dt}=0$.
Find $x$ using Pythagras', the goal is to look for $\cfrac{dk}{dt}$, so with the values you were given $$\cfrac{dx}{dt}=25ft\cdot s^{-1},\ \ k=500ft$$
You should be able to complete it.
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.
Best Answer
Let $R, r$ be the radii of the larger and smaller circles, respectively. Then: $$\pi R^2-\pi r^2=10 \pi \Rightarrow R^2-r^2=10.$$ Differentiate with respect to time $t$: $$2RR'-2rr'=0 \Rightarrow R'=\frac{rr'}{R}=\frac{5\cdot 1}{\sqrt{10+5^2}}=\frac{\sqrt{35}}{7}. $$