1. An airplane, flying horizontally at an altitude of 1 mile, passes directly over an observer. If the constant speed of the airplane is 400 miles per hour, how fast is its distance from the observer increasing 45 seconds later? Hint: Note that in 45 seconds, the airplane goes 5 miles.

  2. A metal disk expands during heating. If its radius increases at the rate of 0.02 inch per second, how fast is the area of one of its faces increasing when its radius is 8.1 inches?

totally don't know the idea in these questions, help, please!


This smells like homework ( ;-) ), so I'm just going to get you set up. If you need more help, leave a comment.

The first problem is best accompanied by a picture. The second doesn't really need one, as the picture is trivial. I'll address the second problem first, as it is somewhat easier.

The Second Problem

The rate of change of the radius is $0.02\frac{in}{s}$. Thus: $$\frac{dr}{dt} = 0.02$$ Area of circle is easy: $$A = \pi r^2$$ Differentiate area: $$\frac{dA}{dt} = \pi((2r)\frac{dr}{dt})$$ All that remains is to plug in the radius and rate of change of radius.

The First Problem

Picture (If the picture doesn't show up, let me know... imgur may have been blocked on my computer.)

$x$ is the horizontal distance the plane has traveled since directly passing over the observer. $d$ is the distance you are seeking. (oops--just realized $d$ won't work as a symbol very well for obvious reasons. I'm going to substitute $r$.) $$\frac{dx}{dt} = velocity = 400$$

What is the distance? $$r = \sqrt{x^2 + 1^2}$$ $$\frac{dr}{dt} = \frac{(2x)\frac{dx}{dt}}{2\sqrt{x^2 + 1}}$$

Now you just have to plug in the values for $x$ and $\frac{dx}{dt}$.

  • $\begingroup$ It's my homework but you know, not just two. The problems is solved, thanks so much! $\endgroup$ – PlusA Nov 25 '12 at 17:25

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.

  • $\begingroup$ @PlusA: You will find the immediate differentiation idea useful. It is easier to differentiate when you don't have ugly square roots around. $\endgroup$ – André Nicolas Nov 25 '12 at 17:37

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