The air drag force experienced by an airplane in flight, Fd,...

The air drag force experienced by an airplane in flight, $F_d$, can be represented by the formula $F_d = \frac{1}{2}CpAv^2$, where $C$ is the drag coefficient; $\rho$ is the air density; $A$ is the cross-sectional area of the plane perpendicular to the direction of flight; and $v$ is the speed of the airplane. If the speed is doubled at a given altitude, what is the resulting increase in the power requirement for the engines. a. increased by a factor of $\sqrt{2}$ b. increased by a factor of 2 c. increased by a factor of 4 d. increased by a factor of 8

Transcript

00:01 One of the first ways that people use to learn calculus is to solve so -called max -min problems.

00:08 The idea is that you have a function of a single variable.

00:13 We'll just call it function of x, but it could be any function with any single variable.

00:19 And if you take the derivative of that function with respect to its variable, that gives you the slope of the tangent line to the curve.

00:34 And what we know about that is that slope is identically equal to zero at a so -called, sometimes called critical points or extreme, it's another term for them, or another term is just max minns.

00:59 It's my personal favorite by the way max minns.

01:02 It's very descriptive.

01:04 So this is a way to find the maximum or the minimum of a function other than plotting it, which certainly is easy enough to do, but much nicer to use an analytical tool.

01:18 With the analytical tools, you can also take a second derivative and evaluate it at that critical point.

01:28 The second derivative of f of x with respect to x gives you the curvature.

01:38 And if you you evaluate that at the critical point, if that curvature is positive, you know you have something curved up and you have found a minimum.

01:49 If it's negative, your curvature is downwards in the function and you found a maximum.

01:55 If the second derivative is zero, by the way, you have found a so -called inflection point where the curvature is changing from one sign to the other.

02:08 And those could be of interest as well.

02:12 So let's take a look at an example.

02:16 So in this example, we are looking at the force of air drag on an airplane that's flying horizontally.

02:29 And it is supposedly equal to two terms, a term that is proportional to the speed squared of the airplane.

02:41 The speed is in the extraction.

02:44 The velocity is in the extraction.

02:46 And here we're just looking at the magnitude of that velocity.

02:51 So it's a sum of two terms.

02:55 And what we know is the force of the engine.

02:57 So there's airplane lift that's canceling the weight of the airplane.

03:02 And so there's a lift on the wings that comes from airflow over the wings.

03:12 We won't get into that, but there is a drag force backwards, and in steady flight, the little propeller must create a engine force forwards.

03:31 So the engine force has to be equal to the force of the drag in the opposite direction, but we're not too worried about that.

03:41 In terms of its magnitude, it must be alpha v squared, plus the other coefficient beta over v squared.

03:49 And what we would like to ask is, is there a speed v that minimizes the engine force needed to maintain the horizontal flight? so what we have is basically a function of a single variable, and we may use the calculus tools.

04:33 So we want to take the derivative of that single variable function with respect to its variable, and set that equal to zero to find all critical values of the speed.

04:54 And so we'll do that.

04:55 We have a polynomial.

04:57 So the first term gives us 2 alpha v, and that's a critical v.

05:05 Minus 2, so we have v to the minus 2 power.

05:10 Minus 2 comes out and gives us v to the minus 3 down in the denominator.

05:19 And so now we want to solve this equation to find the critical velocity, and that's fairly straightforward...

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