this is an exploration in my textbook that we are assigned to do. it basically means we are supposed to use our graphing calculator to help. my calculator won't work for some reason so I'm trying to do it without and cannot figure it out. the question is really irrelevant to what we are learning as well so if anyone could help me with understanding it, that'd be really helpful. thanks ahead of time.
Question: A 10-foot ladder leans against a vertical wall. The base of the ladder is pulled away from the wall at a constant rate of 2 ft/sec.
1. Explain why the motion of the two ends of the ladder can be represented by the parametric equations given on the next page.
X1T=2T
Y1T=0
X2T=0
Y2T=√(10^2 - (2T)^2)
2. What minimum and maximum values of T make sense in this problem?
3. Put your grapher in parametric and simultaneous modes. Enter your parametric equations and change the graphing style to "0" (the little ball) if your grapher has this feature. Set Tmin=0, Tmax=5, Tstep=5/20, Xmin=-1, Xmax=17, Xscl=0, Ymin=-1, Ymax=11, and Yscl=0. You can speed up the action by making the denominator in the Tstep smaller or slow it down by making it larger.
4. Press GRAPH and watch the two ends of the ladder move as time changes. Do both ends seem to move at a constant rate?
5. To see the simulation again, enter "ClrDraw" from the DRAW menu.
6. If y represents the vertical height at the top of the ladder and x the distance of the bottom from the wall, relate y and x and find dy/dt in terms of x and y. (Remember that dx/dt=2.)
7. Find dy/dt when t=3 and interpret its meaning. Why is it negative?
8. In theory, how fast is the top of the ladder moving as it hits the ground?
Thank you so much for anyone that can help. I know it's a lot, I just don't have a working calculator at the moment and there is not much I can do..I'll get a 0 for the entire assignment if I don't have this portion complete so I'm quite desperate, unfortunately. Not the nicest or most considerate of teachers :(
I'd say that (X1,Y1) is the position of the base of the ladder, if the base of the wall is (0,0). Since it is moving at the rate of 2 ft/s, after t seconds it is 2t feet from the wall.
Naturally, the Pythagorean theorem says that the height of the top of the ladder obeys h^2+x^2 = 10^2, so you can see how (X2,Y2) is the position of the top of the ladder.
#2. Clearly, the top cannot be more than ten feet up, nor the base farther than 10 ft away.
#4. Naturally, the longer the base moves away, the faster the top will fall.
#6. since x^2+y^2 = 100
2x dx/dt + 2y dy/dt = 0
#7 should be easy.
#8. In theory, since dy/dt = -x/y, when y=0, the ladder is falling infinitely fast. If you have ever stood on such a falling plank, it sure feels like it!
There are lots of online graphing calculators. wolframalpha, desmos, and many others.