Having trouble finding downward acceleration of top of ladder in a standard, but tricky, ladder and wall problem

Question

An 18-foot ladder extends over a 12-foot wall; the bottom of the ladder is pulled away from the wall at 2'/sec; find the vertical velocity (then acceleration) of the top of ladder when the angle the bottom of the ladder makes with the floor = pi/3 (60 degrees.) It's the acceleration, or second derivative of the 'y' value, that I'm having trouble with.

I posted this a few days ago with my attempted solution to the easier part, the downward velocity of the top of the ladder at the time in question.

I had apologized in advance for any errors ; now I'm apologizing retroactively as well. My solution was incorrect. I had failed to realize that though the length of the ladder itself was a constant (18'), the distance protruding over the top of the wall was changing, thus a variable (call it 'd') which needed to be accounted for.

(I am not sure how to upload a diagram, which would have been helpful)

Using the Pythagorean Theorem, some basic trig to find 'x' (the distance between the foot of the ladder and the wall) and '18-d' (the length of the ladder minus 'd') and two equations, I was able to derive the correct answer, which my text provided, but did not explain.

These are equations I used:

$ 12^2 + x^2 \ = \ (18-d)^2 \ \ \text{and} \ \ \frac{y}{d} \ = \ \frac{12}{18-d} \ \ . $

I found dd/dt to be 1, substituted it back in to the differentiation of the second equation, and derived the textbooks answer of dy/dt = - 1.125 or -9/8. So much for the velocity. Now, for the second part of the question, I'm stumped.

Does anyone know how to use these equations (or perhaps some other equation) to find y'', the downward acceleration? Many thanks, as I'm trying to learn this on my own

(Note: the text in question is Calculus (6th edition) by Varberg and Purcell, 1992; P 155, Q 25.

Answer 1

If you had written down the related-rates equation for $ \ \frac{dy}{dt} \ $ versus $ \ \frac{dx}{dt} \ \ , $ it would have been quicker to proceed from there. We can reconstruct it easily enough, however.

The "overhang" $ \ D \ $ of the ladder beyond the top of the wall forms a similar triangle to the rest of the ladder, so you obtained the equations $ \ 12^2 + x^2 \ = \ (18-D)^2 \ \ , \ \ 12·D \ = \ y·(18 - D) \ \ , $ from which implicit differentiation produces $$ x· \frac{dx}{dt} \ \ = \ \ -(18 - D)·\frac{dD}{dt} \ \ , \ \ 12·\frac{dD}{dt} \ \ = \ \ (18 - D)·\frac{dy}{dt} \ - \ y·\frac{dD}{dt} $$ $$ \Rightarrow \ \ \frac{dD}{dt} \ \ = \ \ \frac{18 - D}{12 + y}· \frac{dy}{dt} \ \ . $$ We would like to "be rid of" the rate $ \ \frac{dD}{dt} \ \ , $ so substitution gives us $$ x· \frac{dx}{dt} \ \ = \ \ - \frac{(18 - D)^2}{12 + y}· \frac{dy}{dt} \ \ = \ \ - \frac{12^2 \ + \ x^2}{12 + y}· \frac{dy}{dt} $$ by application of our "Pythagorean" equation. We now need the current values of $ \ x \ $ and $ \ y \ \ : \ $ as we are given that the ladder is inclined $ \ 60º \ $ to the ground at the moment in question, we have $ \tan 60º \ = \ \sqrt3 \ = \ \frac{12}{x} \ \Rightarrow \ x \ = \ 4·\sqrt3 \ \Rightarrow \ 18 - D \ = \ \sqrt{12^2 + (4 \sqrt3)^2} \ = \ \sqrt{192} \ = \ 8·\sqrt3 $ $$ \Rightarrow \ \ y \ \ = \ \ \frac{12·(18 - 8 \sqrt3)}{8 \sqrt3} \ \ = \ \ 9 \sqrt3 \ - \ 12 \ \ \approx \ \ 3.59 \ \ . $$ Our related-rates equation then yields $$ (4 \sqrt3) \ · \ 2 \ \ = \ \ - \frac{192}{12 \ + \ (9 \sqrt3 \ - \ 12)}· \frac{dy}{dt} $$ $$ \Rightarrow \ \ \frac{dy}{dt} \ = \ - \frac{(4 \sqrt3) \ · \ 2 \ · \ (9 \sqrt3)}{192} \ \ = \ \ -\frac{216}{192} \ \ = \ \ -\frac98 \ \ . $$ So we are able to confirm your results to this point.

For the accelerations, we need to differentiate the related-velocities equation with respect to time. For that, it will be convenient to re-arrange it as $$ \frac{d}{dt} \ [ \ x·(12 + y)· \frac{dx}{dt} \ ] \ \ = \ \ - \frac{d}{dt} \ [ \ (12^2 \ + \ x^2)· \frac{dy}{dt} \ ] $$ $$ \Rightarrow \ \ \frac{dx}{dt} ·(12 + y)· \frac{dx}{dt} \ + \ x·\frac{dy}{dt}· \frac{dx}{dt} \ + \ x·(12 + y)· \frac{d^2x}{dt^2} $$ $$ = \ \ - \ [ \ 2x · \frac{dx}{dt} · \frac{dy}{dt} \ + \ (12^2 \ + \ x^2)· \frac{d^2y}{dt^2} \ ] \ \ , $$ after extensive use of the Product and Chain Rules. The problem statement suggests (it doesn't say so specifically) that $ \ \frac{dx}{dt} \ = \ +2 \ $ is constant, making $ \ \frac{d^2x}{dt^2} \ = \ 0 \ \ . $ Our work tells us that $ \ \frac{dy}{dt} \ = \ -\frac98 \ \ , $ $ x \ = \ 4·\sqrt3 \ \ , \ \ 12^2 \ + \ x^2 \ = \ 192 \ \ , \ \ \text{and} \ \ 12 + y \ = \ 9 \sqrt3 \ \ . $ So we may write $$ (9 \sqrt3)· (2)^2 \ + \ (4 \sqrt3)·\left(-\frac98 \right)· 2 \ + \ 0 \ \ = \ \ - \left[ \ 2·(4 \sqrt3)· 2 · \left(-\frac98 \right) \ + \ 192 · \frac{d^2y}{dt^2} \ \right] $$ $$ \Rightarrow \ \ 36 \sqrt3 \ - \ 9 \sqrt3 \ \ = \ \ 18 \sqrt3 \ - \ 192 · \frac{d^2y}{dt^2} $$ $$ \Rightarrow \ \ \frac{d^2y}{dt^2} \ \ = \ \ -\frac{9 \sqrt3}{192} \ \ = \ \ -\frac{3 \sqrt3}{64} \ \ \approx \ \ -0.0812 \ \frac{\text{ft.}^2}{\text{sec.}} \ \ . $$ So the top of the ladder is undergoing a very small downward acceleration, largely owing to the fact that the angle the ladder makes to the ground is decreasing.