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Handout Monday 10/13
Suppose a 10 foot ladder is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a constant rate. What can we say about how fast the top of the ladder is sliding down the wall?
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Let \(x\) be the distance from the wall to the bottom of the ladder, and let \(y\) be the height of the ladder on the wall.
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We know that \(x^2 + y^2 = 10^2\text{.}\)
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How much does the top of the ladder move when the bottom of the ladder moves 1 foot (from 0 feet to 1 foot)? What if the ladder moved a foot from 5 feet to 6 feet?
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What we want to know is what \(\frac{dy}{dx}\) is for any value of \(x\text{.}\) Now if we had a function for \(y\) in terms of \(x\text{,}\) we would just take the derivative of that function.
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Certainly \(y\) depends on \(x\text{,}\) but we don’t have an explicit function for \(y\) in terms of \(x\text{.}\) We have an implicit relationship between \(x\) and \(y\text{.}\)
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Luckily, we can still compute \(\frac{dy}{dx}\) by differentiating both sides of the equation \(x^2 + y^2 = 10^2\) with respect to \(x\text{.}\) When we get to the term involving \(y\text{,}\) we will need to use the chain rule, since \(y\) is a function of \(x\) (so it is an “inside function”).
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Do this.
There are lots of interesting implicitly defined functions. For example, the equation \(x^3 + y^3 = 9xy\) defines \(y\) implicitly as a function of \(x\text{.}\) (This is called a folium of Descartes.) Graph this using Desmos. Then use implicit differentiation to find \(\frac{dy}{dx}\text{.}\)
Play around with different implicit equations in Desmos. Find one and verify where the tangent really is vertical using implicit differentiation.
