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Calculus sucks... (Homework help.)
Shiekahn_boy
Registered User regular
Registered User regular
So yeah, i'm taking Calculus AP right now in highschool
It's hard... The first day we get homework and i don't know what to do with it.
Here's the problem:
Phoebe Small sits atop a swimming pool slide at t=0. As she slides, Calvin ascertains her speed to be v(t)=10sin(.3t) Feet per second.
Ok, here's what i got from this. Her maximum speed is about 10 feet per second at 300 seconds. Then it slows. But i don't know the variable that tells me the Height, or Length of the bottom of the ladder to the edge of the water. (I'm thinking it;s 10, but I just got back into calculus form last years pre-cal so I'm not sure what I'm doing. Way to throw me in the lake guh...)
I know that the formula above calculates the maximum velocity. What i don't know is the time she hit's the water.
The first question asks:
Plot the graph of Phoebe's speed until she slides into the water. Use a full sheet of graph paper, making the graph large. Determine a suitable practical domain and range for v(t) in this situation. Identify the points when she is at the top of the slide and when she hits the water.
In the beginning it explains that at T (representing x) on the calculator is zero. So when i check it it says that her speed is zero (which makes sense, she isn't going anywhere.) I guess what I'm asking for though is Number that represents the distance she is from the ground? I'm not sure.
It's hard... The first day we get homework and i don't know what to do with it.
Here's the problem:
Phoebe Small sits atop a swimming pool slide at t=0. As she slides, Calvin ascertains her speed to be v(t)=10sin(.3t) Feet per second.
Ok, here's what i got from this. Her maximum speed is about 10 feet per second at 300 seconds. Then it slows. But i don't know the variable that tells me the Height, or Length of the bottom of the ladder to the edge of the water. (I'm thinking it;s 10, but I just got back into calculus form last years pre-cal so I'm not sure what I'm doing. Way to throw me in the lake guh...)
I know that the formula above calculates the maximum velocity. What i don't know is the time she hit's the water.
The first question asks:
Plot the graph of Phoebe's speed until she slides into the water. Use a full sheet of graph paper, making the graph large. Determine a suitable practical domain and range for v(t) in this situation. Identify the points when she is at the top of the slide and when she hits the water.
In the beginning it explains that at T (representing x) on the calculator is zero. So when i check it it says that her speed is zero (which makes sense, she isn't going anywhere.) I guess what I'm asking for though is Number that represents the distance she is from the ground? I'm not sure.
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Shiekahn_boy on
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Secondly, you will want to find the maximum velocity on this range. This will be where sin = 1 so you need to find the relevant time to make this so.
From there you can use your basic knowledge of a sin graph to sketch the graph or calculate some in between points if you want to be more precise.
Sorry if I'm confusing, I find it some what hard to explain and its been a year since Calculus.
EDIT: Actually, yea the velocity being zero when it hits the water doesn't make a bunch of sense either, my bad. In that case, yea we need more information. Sorry.
Plot the graph of Phoebe's speed until she slides into the water. Use a full sheet of graph paper, making the graph large. Determine a suitable practical domain and range for v(t) in this situation. Identify the points when she is at the top of the slide and when she hits the water.
It's asking YOU to find a suitable practical domain and range.
It should be clear this is vertical velocity, negative velocity would just mean the slide's sloping back up, no big deal, at least on the surface. But this function for v(t) is periodic. So what's realistic? She can't hit the water while the slide is sloping back up.
Here: http://people.hofstra.edu/steven_r_costenoble/Graf/Graf.html make sure you put 0.3 and call t x to graph it.
So what we've decided is she can't hit the water while the graph is negative, that's fine. Can it go out for more than one period? From t=0 to t=5, her velocity is increasing, she's accelerating! Around t=5 the acceleration dies and the sin function changes slope and begins decreasing, that means she still has positive velocity but is slowing down, in other words the slide is leveling out. At v=0 between t=10 or 11, the slide actually levels out, then it starts climbing again. The slide is going to slope down and up and since velocity is oscillating between the same points, she's passing physically through the same height over and over. But she can only hit the water once. Therefore the suitable domain is anywhere between t=0 and t=~10, not counting 0 and ~10
If you're curious, to find h(t) you'd have to integrate v(t), and obviously you haven't learned that yet. It's a graphing + understanding exercise
DOUBLE Edit: See my below post, this following part is slightly wrong
Edit: So if y'all don't believe me, what we're given is v(t)=dh/dt = 10*isin(.3*t), separate variables and integrate both sides for h= -10/0.3*cos(.3*t) = -33.33*cos(0.3*t)
Confusingly because positive velocity is defined as in the downward direction, the height starts negative and is opposite of what's intuitive. Point is, h=0 at around t=5, so she hits the water the first time around there. However he needs to see it from observing the graph
However, here is how you'd do the problem given the little bit more information.
The derivative of position is velocity
The derivative of velocity if acceleration
The derivative of acceleration is jerk (this is very very rarely used actually)
So take the indefinent integral of her velocity, ignore the constant part that'll pop up, then plug in t = 0. This will give you her position at t = 0.
To get the position at when she hits the water, just find when she hits the water, then plug that into the same indefinent integral.
Basically, we know that sin oscilates between -1 and 1, however, we know because the amplitude of the wave is 10, it'll oscilate between -10 and 10. This means, the MAX her velocity can ever be is 10m/s (or feet/s if you're in America land). And by association her minimum speed is -10m/s, but this is erronious data. (So far, it might come into play, doubt it).
And yeah, I think you did your original graph wrong.
And yeah, you can figure out where she hits the water from the BlochWave's logic. Thus plug that point into the same indefinent integral.
It's not automatically obvious what a suitable domain and range is, and I tried to explain, using the skills he should possess, what is
When it asks for a suitable domain, it's asking YOU how far it should extend on the t axis, and as noted, it needs to be greater than t=0 and less than ~t=10, if that wasn't clear. The range is going to be from v=0 to v=whatever the equation yields for whatever value of t you choose.
In reality the analytic solution would be h(t) = -33.33*cos(0.3*t) + h(0) + 33.33, where h(0) is the initial height (and will be negative because of the sign weirdness)
So I was worried because the actual t of impact would change depending on the initial height (like duh it should) but because of the nature of the velocity function, initial height can't be greater 66.99 feet (33.33*2) or it never hits the water. If it's set to 66.99 then it just hits the water at t=~10.
For the OP, then, I stand by my original answer. It is suitable to say that the impact occurs between t=0 and t=~10 (should be apparent just what the value is from graphing). Realistically it will depend on the initial height. The maximum possible time it can take is about 10 seconds.
Obviously the height can be so low that it hits right after it starts, so saying t>0 is fine.
The other side of that is it hits jusssst as the slide starts to level out, at around t=10ish (when v=0). Any higher and the slide starts to climb before it hits the water. So t<10ish.
determining domain: it is ridiculous to believe that the slide will ever tilt up or ever do anything but increase the velocity, the domain should be 0<t<Pi/2. this means that T= 5Pi/3 or about 5.23 seconds
Determining range: the velocity goes from 0 to 10 ft/s from t = 0 to 5.23 seconds
Determining ratre of acceleration: simply take the derivative of the velocity to get the equation 3cos(.3t) for acceleration.
Determining length: we take the definite integral of velocity from 0 to 5.23s to get the total distance traveled along the slide (which is the length). the length of the slide is therefore 100/3*cos(Pi/2) - 100/3*cos(0) = 0 + 100/3 or 33.33 feet
I just treated it as a simple ODE initial value problem. I could have rewritten it as a definite integral, but in solving for the constant C using the initial value I remove the indefiniteness
If it levels off velocity will decrease towards 0, I stand by my answer
And again, I wonder if he really is supposed to use integration to solve this on his first day of calculus
questions like this always assume ideal conditions such as ignore friction and drag. meaning that if it leveled off, the velocity would remain constant, acceleration would be 0 and position would increase. thats not the case here.
also, this is AP calc, you are supposed to already know some calculus going into it. specifically how to integrate and differentiate. most likely this problem was a feeler to find out how much the students know/remember and which ones know more.
For reasons unknown to humanity I assumed the function was for her vertical velocity (dy/dt as it were).
I blame alcohol. It's like they say: Don't drink and derive! *crickets*
And although it's unlikely that the slide would curve back up, it might, you know to produce a lifting effect at the end?
Anyways, I think the best thing to do is assume that once she hits her first local max, she's hit the water, and just use integration and some plug and play math to find out her position.
He said its his first day, and I'm betting he barely even knows about limits and differentiation, let alone integrals.
See how many books I've read so far in 2010
pre-calc?
regular calc?
See how many books I've read so far in 2010
What I learned:
She enters the water at about 5.23 seconds at the rate of 10 ft per second.
Ok, Derivative? Not sure i learned that term or have forgotten it...
Yes, she put us up to this to determine who are the quitters.
I'm not, I'm going to struggle some, but I'm not going to quit dammit.
Edit: I'm still confused on how to get the Length of the slide. Can someone explain Step by step?
Well... not true, if you don't know how to integrate you can't do this question... But you need to know derivatives to do integration (not entirely true, but very true to the understanding of what you're really doing).
Anyways, here is step by step what to do (I think...)
Take the definate integral ( -33.33Cos(0.3t) )
Now plug in t = 0, this is her INTIAL point.
Now plug in t = 5.23 this is her FINAL point.
Length of the slide will be final - initial.
Or...
Take the definate integral from 0 to 5.23 (5.23 on top, 0 on the bottom).
Either way is the exact same thing.
But if you seriously don't know what derivatives, limits, or integration is, you can't do this problem. Like, I can tell you the formula to use, and how to use it, but you won't understand what you're doing.
Derivative, the rate of change of an equation (or model depending). Think of it as the slope it every part of your graph
Integral, the area underneath your curve. Think of it as your backwards operation to derivatives.
take a small section of your graph (1 box in width, no restriction on height) and figure out what the x length of that section is. then look at your curve and find out what the average value of the velocity in that slice is.
so for x=0 to .1 you would have a time segment of .1 seconds and an average velocity of 0+sin(.03)/2 = .029 feet which means in the first .1 second you have moved distance = rate * time which is .029 * .1 = .0029 feet. you can do this another 53 times to find the average value across the entire graph or you can select a larger segment such as .25 or .5 seconds. I suggest .5 myself. add all of those values up and you will get the distance traveled which should be about 33 feet.
to find the acceleration you are just looking for the slope of the graph of velocity. so my bet here is that you should just take the starting point of 0 and the ending point of 10 ft/s and divide it by the total time which should give you about 2 ft/s*s. you can do the same trick with splitting it up into segments and adding the segments up if you want.
the basic themes are: distance is the area under the curve, acceleration is the slope, domain and range are both just assumptions based on the problem.
Unless you knew the rate of change of velocity was acceleration, and the rate of change of position was velocity, you wouldn't be able to figure this out as well.
well, distance = rate*time is usually taught in algebra.....
the acceleration means that you need to know some physics or science but still, not too far of a stretch.