Physics Help Forum Question regarding rate of change

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 Jun 12th 2010, 07:35 AM #1 Junior Member   Join Date: Apr 2010 Posts: 12 Question regarding rate of change Sorry if this belongs on the math forum, it's been down all day. The question: A 5m ladder is leaning against a vertical wall. Suppose that the bottom of the ladder is being pulled away from the wall at a rate of 1m/s. How fast is the area of the triangle underneath the ladder changing at the instant that the top of the ladder is 4m from the floor? My attempt: A = (1/2)xy dA/dt = dA/dx + dx/dt dA/dt = (1/2)(dx/dt)y + (1/2)(dy/dt)x Sub in y = 4, x = 3 (using trig), dA/dt = 3.5(dy/dt) I'm not sure if I'm doing this correctly, it doesn't seem right that I'd have to solve for dy/dx. The answer is supposed to be 7/8. Any assistance would be great!
 Jun 14th 2010, 03:09 AM #2 Member   Join Date: Jun 2010 Location: CT, USA Posts: 35 The chain rule application was incorrect. However, the line after it was fine. But then, I never saw any use of the Pythagorean theorem, which is necessary to relate x and y. There is one step in which you seem to have regarded dx/dt and dy/dt as like terms?!? I would start the problem as follows: x^2+y^2=25, and hence x(dx/dt)+y(dy/dt)=0. Also, A=(xy)/2, and therefore dA/dt=(x(dy/dt)+y(dx/dt))/2. You know dx/dt, so by virtue of an equation above, you should be able to figure out dy/dt. Once you have x, dx/dt, y, and dy/dt, you just plug into the expression for the derivative of the area, and you're done. I think you had most of the basic ideas correct, just some mechanics of solving it were incorrect. Incidentally, what happens as y->0? __________________ "Tell me and I forget. Show me and I remember. Involve me and I understand." - Confucius "Excite and direct the self-activities of the pupil, and as a rule tell him nothing that he can learn himself." - The Seven Laws of Teaching, by John Milton Gregory
 Oct 7th 2010, 11:02 PM #3 Junior Member   Join Date: Oct 2010 Posts: 5 I believe the reason he teaches it like this is because the class is AP Physics B, which is the non-calculus AP Physics. I guess that teaching it this way somehow makes it easier for the students not sure how, but that's just me, given an object thrown into the air at 10 m/s, I would just graph a curve of its displacement and use its equation and derivatives thereof to find its position/velocity/acceleration at any given time point. Last edited by werehk; Oct 8th 2010 at 01:34 AM.

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