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Handout Monday 10/20
Calculus gives us tools to understand how quantities change. We have seen this appear primarily when we are given a function that describes the values of a quantity in terms of some independent variable. But there are also interesting examples of multiple quantities that change in tandem, both functions of some common variable. These related quantities with have related rates of change.
Let’s start with a classic example: a ladder is sliding down a wall. Previously, we assumed that the bottom of the ladder was moving away from the wall at a constant rate, and described the top of the ladder’s speed in terms of the distance from the wall to the base of the ladder.
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Now let’s assume that we don’t know about the speed of the bottom of the ladder. We just want to describe the relationship between the speed of the top of the ladder and the speed of the bottom of the ladder, both in inches per minute (so in terms of time, not height in terms of distance).
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The relationship between the distance from the wall to the base of the ladder \(x\text{,}\) and the height of the top of the ladder \(y\text{,}\) can be described using the Pythagorean Theorem.
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We have \(x^2 + y^2 = l^2\text{,}\) where \(l\) is the length of the ladder. Both \(x\) and \(y\) are functions of time, \(t\text{.}\) The length of the ladder is presumably constant, let’s say 10 ft.
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Just like we did with implicit differentiation, we will take the derivative of both sides of this relationship, but this time with respect to \(t\) (instead of with respect to \(x\)).
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We get \(2x\frac{dx}{dt} + 2y \frac{dy}{dt} = 0\text{.}\)
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We can now answer lots of questions, provided we know two or three of these four quantities/rates. For example, how fast is the top of the ladder moving if when the base of the ladder is 6 feet from the wall, it is moving at a rate of 3 feet per second? Note that we can determine the value of \(y\) from knowing \(x\text{,}\) using the original (static) relationship (we get \(y = 8\)). Since we also know \(\frac{dx}{dt}\text{,}\) we can substitute three values and get the fourth.
Return to the handout from last class with the other related rates scenarios. Try using implicit differentiation to find the relationship between their rates. Do as many as there is time for (start from the last question, since we didn’t ever write down the original static relationship).
