Relate simultaneous rates before substituting

Try

A rigid 13 metre ladder touches a vertical wall and a level floor. Its foot is 5 metres from the wall, making its top 12 metres high. If the foot moves outward at 2 metres per second, predict the sign of the top’s height rate. Then explain why the height rate need not also have magnitude 2.

Predict

Imagine the ladder almost horizontal, with the top very close to the floor. Predict whether maintaining the same foot speed would demand a gentle or rapid change in height. This is a prediction about the rigid contact geometry, not yet a statement about real forces or falling dynamics.

Experiment

Label the horizontal foot position x(t), vertical top position y(t) and fixed length 13. The Pythagorean relationship is x²+y²=169. Differentiate in time to obtain 2x x′+2y y′=0, then solve y′=−x x′/y.

Instantaneous x	Instantaneous y	Prescribed x′	Computed y′
5	12	2	−5/6
12	5	2	−24/5

Distances are metres and rates metres per second. Both rows satisfy the same length constraint. Sketch the two triangles at the same scale. The change in rate is caused by geometry, even though the prescribed foot speed is unchanged.

Understand

A related-rate problem connects derivatives of quantities linked by a model equation. First define changing variables and fixed parameters. Next write a relationship valid throughout nearby times. Differentiate that relationship with respect to time. Only then substitute the values describing the requested instant.

Replacing x(t) with 5 before differentiating turns a changing quantity into a constant and erases its rate. A value at one instant is not an identity over an interval. The ladder’s length, however, is legitimately constant under the stated rigid-ladder assumption, so its squared derivative contribution is zero.

For a cone-shaped tank of total height 6 metres and radius 3 metres, the changing water radius r and depth h obey r/h=3/6 by similarity. The water volume is V=πr²h/3=πh³/12. If V′=2 cubic metres per minute, differentiating gives 2=(πh²/4)h′. At h=2, the depth rate is 2/π metres per minute. Using the full tank radius 3 for the partially filled water surface would produce the wrong relationship.

A shadow problem has a different distinction. With a 6 metre lamp and a 2 metre person, let x be the person’s distance from the lamp and s the shadow length. Similar triangles give 6/(x+s)=2/s, hence s=x/2. If x′=3 metres per second, the shadow length grows at 3/2, but the tip position x+s moves at 9/2. The requested position determines which rates must be added.

The ladder formula becomes unbounded as y approaches zero while positive x′ remains fixed. It exposes a limit of the stipulated motion. A real ladder may lose contact or violate the prescribed speed, and impact needs a dynamical model. Do not claim that correct differentiation validates every physical assumption.

Apply

In the practice bank, begin with a labeled sketch and a list of known values and rates. State the requested signed rate with units. For the tank, reduce the changing radius using similarity before differentiating. For the ladder model critique, distinguish an algebraic consequence from a physical guarantee. Revisit the circle related-rate lab to compare a derivative with a finite-step change.

Continue

Next use theorems to obtain rates that must occur somewhere in an interval, even when an explicit location is unknown. Their continuity and differentiability hypotheses are essential parts of the argument.