A plane is located x=40 miles (horizontally) away from an...

A plane is located $x=40$ miles (horizontally) away from an airport at an altitude of $h$ miles. Radar at the airport detects that the distance $s(t)$ between the plane and airport is changing at the rate of $s^{\prime}(t)=-240 \mathrm{mph} .$ (a) If the plane flies toward the airport at the constant altitude $h=4,$ what is the speed $\left|x^{\prime}(t)\right|$ of the airplane? (b) Repeat with a height of 6 miles. Based on your answers, how important is it to know the actual height of the airplane?

A plane is located $x=40$ miles (horizontally) away from an airport at an altitude of $h$ miles. Radar at the airport detects that the distance $s(t)$ between the plane and airport is changing at the rate of $s^{\prime}(t)=-240 \mathrm{mph} .$
(a) If the plane flies toward the airport at the constant altitude $h=4,$ what is the speed $\left|x^{\prime}(t)\right|$ of the airplane?
(b) Repeat with a height of 6 miles. Based on your answers, how important is it to know the actual height of the airplane?

Key Concepts

Related Rates

This concept involves finding the rate at which one quantity changes with respect to time in relation to the rate of change of another quantity. In problems like this, changes in distances or positions are interrelated through a known geometric relationship, and the derivative with respect to time (denoted by prime notation) allows us to connect these rates.

Pythagorean Theorem

The Pythagorean Theorem is used to relate the horizontal distance and the constant altitude of the airplane to the straight?line distance from the radar. This relationship, in the form s^2 = x^2 + h^2, is fundamental in setting up the equation whose differentiation leads to finding the unknown speed.

Implicit Differentiation

Implicit differentiation is applied to the Pythagorean equation to differentiate both sides with respect to time. This allows us to connect the rate of change of the distance between the plane and the airport (s') with the rate of change of the horizontal distance (x'), even though x and s are interdependent.

Sensitivity to Model Parameters

The problem illustrates how small variations in one of the parameters (the altitude ‘h’) affect the computed speed of the airplane. It highlights the importance of accurately knowing the parameters in a mathematical model, especially in real-world applications like navigation, even though the impact may be small depending on the relative sizes of the dimensions involved.

Transcript

00:01 For this problem, we're given that we're at an airport and there is a plane 40 miles out horizontally, which is what our blue line indicates here.

00:10 At an unknown height age, then the distance from that airplane to the airport is represented by s of t.

00:22 So that's our given information in conjunction with we know that the distance between the airplane and the airport is decreasing by 240 miles per hour.

00:39 So for part a, we're asked if the height is four, and we are flying at a consistent height, what will x prime of t be? and so if this is x of t, this is h of t, well, here we can see that this is a right triangle, which, using our intuition, we come up with the pythagorean theorem.

01:09 So x squared plus h squared equals s of t squared, just keeping it all in how we've defined it.

01:20 And so first we do need what is.

01:25 So we want to find what is s of t when h is equal to four.

01:34 So first we're just going to plug in our values.

01:38 So we have 40 squared plus we said h is 4 squared equals s of t.

01:46 And we just want s of t by itself.

01:48 So we're going to square root this.

01:50 When you calculate that out, that does come out to 4 square root of 101 is equal to our s of t.

02:01 So like we said before, we know this.

02:05 And our h of t function is a constant.

02:09 So if we look at a graph for whatever constant, and i have a constant line, the derivative of this line is just going to be equal to zero.

02:26 So now we want to use this equation.

02:32 So we have x squared plus h squared is equal to s of t squared.

02:40 So keep in mind that since h of t or h is a constant when we take the derivative with respect to t, so it would be 2h, d, h, d, t, or h prime of t.

02:57 This is zero, so that's just going to cancel out this entire part of our equation.

03:04 So we're just not even going to write it to save some space.

03:08 And so taking the derivative of x squared with respect to t which would be to x and to put it in terms of how they are wanting it which is x prime of t we're going to do times x prime of t equals because we're going to pretty much just cancel out this h squared since the derivative is going to be equal to zero and we're multiplying everything of the h squared by zero.

03:39 So then we have 2 s of t and we're taking it with respect to t.

03:45 So we need s prime of t...

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