Here are two examples I think are interesting.

1. A ladder that is leaning against a wall starts slipping down. If the point where the ladder touches the ground (draw your own picture) is moving away from the wall at a constant rate, is the point where the ladder touches the wall falling at a constant rate? People may reasonably think that because the ladder is "rigid" the answer should be yes.

Differentiating $x^2 + y^2 = L$ (where $L$ is the length of the ladder, a constant) shows $dx/dt$ being constant certainly doesn't make $dy/dt$ constant, and in fact since $dy/dt = -x(dx/dt)/\sqrt{L-x^2}$ we see that as $x$ increases (while being less than $L$) and $dx/dt$ is fixed the point where the ladder touches the wall is dropping faster and faster. Some students may even say that from their experience or physical intuition this actually makes sense, which raises the question of whether this is truly a physical phenomenon or a purely mathematical one that has been revealed from calculus.

2. There is a gas is in a chamber with a flexible wall (so the chamber can expand or contract, e.g., a piston is at one end). According to the chemists, if we maintain the gas at a constant temperature while increasing or decreasing the size of the chamber then the pressure $P$ and volume $V$ satisfy the relation $PV$ = constant. (I am thinking of the ideal gas law $PV = nRT$, where $n$ and $T$ don't change.) One aspect of this equation which is obvious is that as the volume goes up/down, the pressure goes down/up. Question: If we decrease the volume at a constant rate, does the pressure increase at a constant rate, and more precisely at the rate which is the reciprocal of the rate at which the volume is going down? I think it's quite natural for people to make a snap judgement that if $dV/dt = -4$ then $dP/dt = 1/4$ because $PV$ is constant, but of course the product rule shows this is wrong, and moreover if $dV/dt$ is constant then $dP/dt$ is definitely not constant.

It's perhaps worth first discussing a situation where such intuition is right, namely where the sum of two variables is fixed, rather than the product. Find your own example where two variables $x$ and $y$ satisfy $x+y$ = constant. Then $dx/dt = -dy/dt$, which makes a lot of sense: the rate at which one goes up is exactly "opposite" to the rate at which the other goes down. But when the product is constant this is completely incorrect.

Here are two examples I think are interesting. (Update: I have added a third example at the end, more intricate than the other two.)

1. A ladder that is leaning against a wall starts slipping down. If the point where the ladder touches the ground (draw your own picture) is moving away from the wall at a constant rate, is the point where the ladder touches the wall falling at a constant rate? People may reasonably think that because the ladder is "rigid" the answer should be yes.

Differentiating $x^2 + y^2 = L$ (where $L$ is the length of the ladder, a constant) shows $dx/dt$ being constant certainly doesn't make $dy/dt$ constant, and in fact since $dy/dt = -x(dx/dt)/\sqrt{L-x^2}$ we see that as $x$ increases (while being less than $L$) and $dx/dt$ is fixed the point where the ladder touches the wall is dropping faster and faster. Some students may even say that from their experience or physical intuition this actually makes sense, which raises the question of whether this is truly a physical phenomenon or a purely mathematical one that has been revealed from calculus.

2. There is a gas is in a chamber with a flexible wall (so the chamber can expand or contract, e.g., a piston is at one end). According to the chemists, if we maintain the gas at a constant temperature while increasing or decreasing the size of the chamber then the pressure $P$ and volume $V$ satisfy the relation $PV$ = constant. (I am thinking of the ideal gas law $PV = nRT$, where $n$ and $T$ don't change.) One aspect of this equation which is obvious is that as the volume goes up/down, the pressure goes down/up. Question: If we decrease the volume at a constant rate, does the pressure increase at a constant rate, and more precisely at the rate which is the reciprocal of the rate at which the volume is going down? I think it's quite natural for people to make a snap judgement that if $dV/dt = -4$ then $dP/dt = 1/4$ because $PV$ is constant, but of course the product rule shows this is wrong, and moreover if $dV/dt$ is constant then $dP/dt$ is definitely not constant.

It's perhaps worth first discussing a situation where such intuition is right, namely where the sum of two variables is fixed, rather than the product. Find your own example where two variables $x$ and $y$ satisfy $x+y$ = constant. Then $dx/dt = -dy/dt$, which makes a lot of sense: the rate at which one goes up is exactly "opposite" to the rate at which the other goes down. But when the product is constant this is completely incorrect.

3. Relate the rates between angular velocity and linear velocity of a crank-and-piston system. It would be too involved to present the details here, and I don't have a good image to include, so see the YouTube video here or read about it here. I find this far more interesting than the previous two examples, and covered it in a calculus lecture yesterday as the last example.

Here are two examples I think are interesting.

1. A ladder that is leaning against a wall starts slipping down. If the point where the ladder touches the ground (draw your own picture) is moving away from the wall at a constant rate, is the point where the ladder touches the wall falling at a constant rate? People may reasonably think that because the ladder is "rigid" the answer should be yes.

Differentiating $x^2 + y^2 = L$ (where $L$ is the length of the ladder, a constant) shows $dx/dt$ being constant certainly doesn't make $dy/dt$ constant, and in fact since $dy/dt = -x(dx/dt)/\sqrt{L-x^2}$ we see that as $x$ increases (while being less than $L$) and $dx/dt$ is fixed the point where the ladder touches the wall is dropping faster and faster. Some students may even say that from their experience or physical intuition this actually makes sense, which raises the question of whether this is truly a physical phenomenon or a purely mathematical one that has been revealed from calculus.

2. There is a gas is in a chamber with a flexible wall (so the chamber can expand or contract, e.g., a piston is at one end). According to the chemists, if we maintain the gas at a constant temperature while increasing or decreasing the size of the chamber then the pressure $P$ and volume $V$ satisfy the relation $PV$ = constant. (I am thinking of the ideal gas law $PV = nRT$, where $n$ and $T$ don't change.) One aspect of this equation which is obvious is that as the volume goes up/down, the pressure goes down/up. Question: If we decrease the volume at a constant rate, does the pressure increase at a constant rate, and more precisely at the rate which is the reciprocal of the rate at which the volume is going down? I think it's quite natural for people to make a snap judgement that if $dV/dt = -4$ then $dP/dt = 1/4$ because $PV$ is constant, but of course the product rule shows this is wrong, and moreover if $dV/dt$ is constant then $dP/dt$ is definitely not constant.

It's perhaps worth first discussing a situation where such intuition is right, namely where the sum of two variables is fixed, rather than the product. Find your own example where two variables $x$ and $y$ satisfy $x+y$ = constant. Then $dx/dt = -dy/dt$, which makes a lot of sense: the rate at which one goes up is exactly "opposite" to the rate at which the other goes down. But when the product is constant this is completely incorrect.

Here are two examples I think are interesting.

1. A ladder that is leaning against a wall starts slipping down. If the point where the ladder touches the ground (draw your own picture) is moving away from the wall at a constant rate, is the point where the ladder touches the wall falling at a constant rate? People may reasonably think that because the ladder is "rigid" the answer should be yes.

Differentiating $x^2 + y^2 = L$ (where $L$ is the length of the ladder, a constant) shows $dx/dt$ being constant certainly doesn't make $dy/dt$ constant, and in fact since $dy/dt = -x(dx/dt)/\sqrt{L-x^2}$ we see that as $x$ increases (while being less than $L$) and $dx/dt$ is fixed the point where the ladder touches the wall is dropping faster and faster. Some students may even say that from their experience or physical intuition this actually makes sense, which raises the question of whether this is truly a physical phenomenon or a purely mathematical one that has been revealed from calculus.

2. There is a gas is in a chamber with a flexible wall (so the chamber can expand or contract, e.g., a piston is at one end). According to the chemists, if we maintain the gas at a constant temperature while increasing or decreasing the size of the chamber then the pressure $P$ and volume $V$ satisfy the relation $PV$ = constant. (I am thinking of the ideal gas law $PV = nRT$, where $n$ and $T$ don't change.) One aspect of this equation which is obvious is that as the volume goes up/down, the pressure goes down/up. Question: If we decrease the volume at a constant rate, does the pressure increase at a constant rate, and more precisely at the rate which is the reciprocal of the rate at which the volume is going down? I think it's quite natural for people to make a snap judgement that if $dV/dt = -4$ then $dP/dt = 1/4$ because $PV$ is constant, but of course the product rule shows this is wrong, and moreover if $dV/dt$ is constant then $dP/dt$ is definitely not constant.

It's perhaps worth first discussing a situation where such intuition is right, namely where the sum of two variables is fixed, rather than the product. Find your own example where two variables $x$ and $y$ satisfy $x+y$ = constant. Then $dx/dt = -dy/dt$, which makes a lot of sense: the rate at which one goes up is exactly "opposite" to the rate at which the other goes down. But when the product is constant this is completely incorrect.