Thread: Generic Calculus Questions

Originally Posted by Invader

An electron moving along the x axis has a position given by x = 12te^(-2.1 t) m, where t is in seconds. How far is the electron from the origin when it momentarily stops?



The answer is 2.102m, but I only know that because I understand that I need to find the derivative of x = 0 for t (x's slope will be 0 when it 'stops moving'). Time t will give me the position in the original equation. The actual values I got only because I know a thing or two about getting around my calculator, but I want to know how to take the derivative of the original equation on paper. Can anyone show me how to do this?


[EDIT]: dx/dt = (12 - 25.2t)e^(-2.1t) is that correct? If so, then my next problem involves getting t onto one side of the equation so that I can solve for it.

You shouldn't have to solve for t here if you only want the zeros. 0 is special because if f(x)y(x) = 0 then you can be sure that f(x)=0 or g(x) = 0. so you need to find the zeros of f(x) = 12 - 25.2t and g(x) = exp(-2.1t)

g doesn't have any roots and f is linear so it only has one which is 12/25.2 to whatever accuracy you have. I never can remember the significant digits thing.

That would be a very hard problem to solve if you needed to get a function for t. You'd just want to use newtons method probably.

Heh, that's right, I could have done it that way. Though I think you misread the
constant e in the original equation as exp^ (there is an exponent in
the original too, don't get me wrong).

I'll show you the way I see this kind of problem, and why I went about it the
way I did. I drew a graph for your viewing pleasure.



You can see that where X is zero in the derivative, I get a t to plug into my
main equation that gives me the value of X that I'm looking for (where the
slope is zero). Your way would have been faster right off the bat, but I have a
habit of seeing the problem laid out like the graph above, except that I don't
draw it out.

(should probably point out that I only used the green line to match up the
f(t) and dx/dt appropriately)

That's a really good way to look at it. As the functions get more difficult though, one needs to learn tricks too But keep looking at it geometrically as long as you can. Going between the two without having to rely on either is key to being good at math as it gets more advanced.

BTW, exp(x) = e^x but I don't have to look around for the carrot if I'm feeling lazy. It gets used if you need to put something big in the exponent as well. You see it around more and more as you deal with bigger equations. Try sticking an integral in the exponent

Originally Posted by PhilosopherStoned

BTW, exp(x) = e^x but I don't have to look around for the carrot if I'm feeling lazy.

See, that was never explained to me, even through calc 1. I always assumed
exp was equivalent to the carrot, in that it meant "exponent". Good Game.
I'll have to pass on the integral in the exponent for now, but I'm sure I'm going
to need a decent explanation eventually.

On that note, I'll be using this thread as I encounter new problems I have
difficulty with.

Thank you for volunteering to be my tutor.

No problem. I love this stuff. I could use a review on all this stuff anyway. I quit math for years so I'm rusty on all levels. I've been working on it again for the past month or so though so I've got a lot of the kinks worked back out.

Yeah just treat exp(x) as a constant and get rid of it, it's never 0.

it's more general than treating one of them as a constant though. it works any way that you can factor a function. take sin^5(x)cos^2(x)x^2/(x^2 + 1).

We can factor this into sin^5, cos^2 and y=(x^2)/(x^2 + 1).

then the whole function can only be zero where at least one of those three is.

It's very nice for finding zeros of a function to not have to invert it.

tbh I don't know of any other ways to solve an equation other than setting each factor to zero. What's inversion in this context?

if f is a function, g is its inverse if f(g(x)) = x. so the inverse of y = 2x is x = 1/2y. It's essentially expressing x as a function of y. We can only do it globally when dy/dx is never zero. If it's zero somewhere, we have to break it into pieces on either side of where it's zero. for example, if y = x^2, then dy/dx = 2x which is 0 when x=0. So we have one inverse of y on either side of 0, these are x = sqrt(y) for x > 0 and x = -sqrt(y) for x > 0.

topologically, whats going on is that the function y = x^2 provides a "double cover" of the positive side of the y axis. you can see this in the graph because every positive value of y can be gotten to by two distinct values of x, namely x and -x. We have to pull it back apart to invert the y=x^2.

I've been trying to figure this out for a grand total of 4 hours now. My book
doesn't explain how to figure two possible angles in a problem like this. My
professor hasn't come anywhere near explaining this yet. Most of the
homework problems have involved learning the new concepts for myself.
Instead of going into a rant, though, I'd just rather be done with this problem
before I blow a fuse out of frustration.

Alright. I think that discovering the possible values of t for both the
horizontal and vertical components of the velocity are going to crucial to
solving for θ. I have no idea anymore.
We know the horizontal distance, and can figure t to be (horizontal
distance)/(velocity of the x component). That would give us
t = 46/32cosθ or 1.44/cosθ
But t is also dependent on the rise and fall of object (vertical velocity
component) at whatever θ is.

I feel like I'm going about this all wrong. I know that I'm supposed to be
solving for two values of θ, and I know that the position function can be
solved with a quadratic to give me two values of t, which means I can figure
for two separate x-velocity and y-velocity components, which in turn would
give me those two angles. Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaah.....

There are two angles that will hit the goal. These angles bounds of the range that you are looking for. So you can solve them one at a time. Do you know how to solve those problems?

Also, the x velocity component is constant at v0*cosθ and the y velocity component is just t*v0*sinθ

I think I need the two angles that are going to hit the goal 3.44m above the ground. And I don't know how to solve for them individually. I was given example problems in place of this question, both of which I solved quickly, but they weren't exactly the same as what I'm being presented with here.

Resolve the initial velocity vector (u) and hence get equations for the horizontal and vertical displacement:

x = ucos@t
y = usin@t -g/2*t^2

Eliminate t. Convert the sec to a tan. Plug in the x and y values. Solve resultant quad and trig equations.

yeah, that's it xei (except you forgot the part where you integrate to get the equations you wrote down ). Fuck I hate trig equations....

Some of them are funky. IIRC cos@ + cos2@ + cos 3@ = 0 is quite tricky.

Got @ = 17.59, 76.69 deg btw, but it's a shit question. Probs slipped up somewhere.

0 = -1.01tan^2(θ) + 46tanθ -4.05

yields 5.08 and 88.74 degrees, but it's incorrect. You have the right answer, Xei, I'm doing something wrong.

I'd recommend you use x, y, u, and @, until you have obtained the general projectile equation. Post it up here and I'll check. Then it's just a matter of setting x = 46, y = 3.44, u = 32 and solving for tan@.

Apologies for the late response. I'll show you how I've been trying to work this out, and you can point out where I'm making my blunder.

u= 32m/s
x= 46m
y= 3.44m

x = ucos@t
y = usin@t -g/2*t^2


solving for x we get:
46 = 32cosθt
t = 1.438secθ

Plugging in t for Y:
3.44 = 32sinθ(1.438secθ) - 4.9(1.438secθ)^2
0 = -10.132(1 + tan^2(θ)) + 46.02tanθ - 3.44.........<----Where I'm messing up, I think.

And, of course, solving for tanθ with the quadratic gives me the wrong answer.

There's way too many numbers in there. They tend to confuse the situation. Do what Xei said and just use letters. That way, it's much easier to spot mistakes.

That way, t = (x/u)secθ

0 = u(sinθ)(x/u)secθ - g/2[(x/u)secθ]² - y
= xtanθ - (gx²/2u²)sec²θ - y
= xtanθ - (gx²/2u²)(1 + tan²θ) - y
= -(gx²/2u²)tan²θ - xtanθ - [y + (gx²/2u²)]

That looks right but I'm more than little stoned so check it and make sure .

At this point, you can 'plug and chug'

I think that last term should look more like:

+ [-y + (gx^2/2u^2)] (How are you entering the superscripts?)
Also, xtanθ is positive. In line 4 you switched the sign.

But other than that it looks a lot easier on the eyes now. I'll be plugging in in just a moment.

-(gx²/2u²)tan²θ + xtanθ - [y + (gx²/2u²)]

is right. I did mess up on the sign of tanθ but the constant term is - [y + (gx²/2u²)]
because -(gx²/2u²) distributes over (1 + tan²θ)

I just use the sup and /sup tags in square brackets. I think it works for sub and /sub but haven't tried it yet.

"because -(gx²/2u²) distributes over (1 + tan²θ)"

That's right, my bad. And here I'm wondering why wrong answers continue to
plague me. God forbid I ever work for NASA. Plugging in again.

I believe you should be aiming for

y = xtan@ - gx^2(tan^2@ + 1)/2u^2