I have to take the FE in April and haven't had a math class since 2001, so can someone please jog my memory on this one:

 

How fast is the distance of an airplane from a fort increasing 1 minute after passing directly over the forst at a height of 10,000 feet when it is flying horizontally at the rate of 200ft/sec?

 

I used the right triangle with side a = 10,000 ft and side b = 12,000 ft (from 60 sec * 200 ft/sec). C is the distance of the plane from the fort.

c = sqrt(a^2 * b^2). I solved for dc/db, is this correct? I ended up with:

dc/db = b

---------------- .

sqrt(a^2 + b^2)

If I plug and chug from here, I don't get the answer given in the book. However, if I use b^2 in the numerator, then I get the right answer, but how does the numerator become b^2?

Do you know how to use relative motion? Suppose an observor is standing in the fort and is watching the whole scenario, a triangle will form.

 

[math]r_A + r_{B/A} = r_B[/math] where r is the distance.

[math]r_A = 10,000 ft[/math]

[math]r_{B/A} \rightarrow d=vt \rightarrow 12,000ft[/math]

 

Use Pythragroas Theroem, you get [math]15,620.5ft[/math]

 

From the [math]r_A + r_{B/A} = r_B[/math] formulae, a new equation is formed: [math]v_A + v_{B/A} = v_B[/math], where v is the velocity.

 

Since the airplane went at 200ft/sec, you have:

[math]v_A + 200ft/sec = v_B[/math].

 

I'm not sure what to do next.

 

But if you want the derivative of [math]\sqrt{a^2+b^2}[/math], it is [math]\frac{2a2b}{2 \sqrt{a^2+b^2}}[/math].

I solved for dc/db, is this correct?

 

No, that is not correct. That is, [math]dc/db[/math] is not the rate at which the distance to the plane is changing. You did calculate [math]dc/db[/math] correctly.

 

[math]dc/db[/math] is unitless. You want something whose units are length/time. In particular, you want [math]dc/dt[/math]. You need to use the chain rule for that:

 

[math]\frac{dc}{dt} = \frac{dc}{db}\frac{db}{dt}[/math]

 

Applying [math]a = 10000\,\text{ft},\ b = 12000\,\text{ft}, \dot b = 200\,\frac{\text{ft}}{\text{sec}}[/math] yields

[math]\frac{dc}{dt} = 200 \frac{12000}{\sqrt{10000^2+ 12000^2}} = 153.64\,\frac{\text{ft}}{\text{sec}}[/math]

 

But if you want the derivative of [math]\sqrt{a^2+b^2}[/math], it is [math]\frac{2a2b}{2 \sqrt{a^2+b^2}}[/math].

 

This is wrong. Look at the units. Verifying that a computation has the correct units provides a good sanity check. The result of any computation is definitely wrong if the units are wrong. (The opposite is not true: Just because the units are right does not mean the result is correct.)

 

[math]\frac{dc}{db}[/math] should be unitless. You answer has units of length.

alternatively yo can use a more direct approach and create a formula for c(t) because the length of side b is vt (velocity x time) you can compute c as

 

(a^2+(v^2)(t^2))^(1/2)

 

differentiating this with respect to time you should get

 

(v^2)t/c

 

where c is the formula given above.

 

Its essentially the same thing as D H's solution, but the relationship to time is more apparent.

[math]\frac{dc}{dt} = \frac{dc}{db}\frac{db}{dt}[/math]

 

Applying [math]a = 10000\,\text{ft},\ b = 12000\,\text{ft}, \dot b = 200\,\frac{\text{ft}}{\text{sec}}[/math] yields

[math]\frac{dc}{dt} = 200 \frac{12000}{\sqrt{10000^2+ 12000^2}} = 153.64\,\frac{\text{ft}}{\text{sec}}[/math]

 

I'm not sure how exactly you came up with the formulae of [math]200 \frac{12000}{\sqrt{10000^2+ 12000^2}}[/math]?

I'm not sure how exactly you came up with the formulae of [math]200 \frac{12000}{\sqrt{10000^2+ 12000^2}}[/math]?

 

[math]\frac{db}{dt} = 200\,\frac{\text{ft}}{\text{sec}}[/math],

one of the given conditions.

 

[math]\frac{dc}{db} = \frac{12000}{\sqrt{10000^2+ 12000^2}}[/math],

from differentiating [math]c^2 = a^2 + b^2[/math] with respect to [math]\frac{dc}{db} = \frac{b}{c} = \frac{12000}{\sqrt{10000^2+ 12000^2}}[/math]

 

[math]\frac{dc}{dt} = \frac{dc}{db}\,\frac{db}{dt} = \frac{12000}{\sqrt{10000^2+ 12000^2}}\,200 = 200\,\frac{12000}{\sqrt{10000^2+ 12000^2}}[/math],

from the chain rule, applying the above, and rearranging.

 

As CPL. Luke noted, another way to solve this is to differentiate [math]c^2 = a^2 + b^2[/math] with respect to [math]t[/math] directly:[math]2c\frac{dc}{dt} = 2b\frac{db}{dt}[/math], or

[math]\frac{dc}{dt} = \frac{b}{c}\,\frac{db}{dt}[/math]

153.64 ft/sec is the correct answer. Thank you very much for the help guys.