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Physics algebra and calculus help

  • Thread starter brusier
  • Start date
  • #1
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Homework Statement



An electron moving along the x axis has a position given by x = 18te^-1t m, where t is in seconds. How far is the electron from the origin when it momentarily stops

Homework Equations


Obviously v(t) = x'(t) so differientiation of x(t) is necessary to find v(t)=0.
From this, we can plug t into x(t) to calculate the position of the electron when the particle momentarily stops [v(t)=0].


The Attempt at a Solution


However, I am unfamiliar in dealing with vairables in exponents.

I tried to calculate x'(t) and got: -18t^2 e^-2t.

I'm pretty sure this calculation is wrong and if its not, I'm very sure I don't know how to solve x'(t) = 0!!
help?

thanks
 

Answers and Replies

  • #2
for x=e^at

dx/dt = ae^at

and for x =

using the product rule, x = f(t)g(t)

dx/dt = f(dg/dt) + g(df/dt)


i get something like

18m(1-t)e^-t
 
  • #3
LowlyPion
Homework Helper
3,090
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Won't you want to use the quotient rule for differentiation?

(f / g)' = (f'*g - f*g') / g2

18*t*e-t = 18*t /et

(18*t*et- 18*et) / e2t

When you set it to 0 looks like a lot of stuff disappears.

18 * t = 18
 
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  • #4
doesn't 18t go to 18 ? the integral of 18t is 9t2

using the quotient rule...
m(18*et - 18t*et)/e2t = 18*m*e-t(1-t)
 
Last edited:
  • #5
LowlyPion
Homework Helper
3,090
4
doesn't 18t go to 18 ? the integral of 18t is 9t2

using the quotient rule...
m(18*et - 18t*et)/e2t = 18*m*e-t(1-t)
Yes of course. 18t and 18. I flipped my integration and derivative. Sorry.

So t = 1.
 
  • #6
27
0
...don't know what m is... but I did try the quotient rule...and got different roots in the second term

x(t) = 18t/e^-t

x'(t) = e^t*18 - 18t*-te^(-t-1)/e^2t

simplyify:

18e^t +18t^2*e^(-t-1) /e^2t

further:

18[e^t - t^2*e^(-t-1)]/e^2t

finally:

18e^(-t-1) (e^(2t+1) - t^2)/e^2t

i'd love to get rid of the denom but I'm sure I did any of this correctly.
 
  • #7
20
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...don't know what m is... but I did try the quotient rule...and got different roots in the second term

x(t) = 18t/e^-t

...
From your original question, you said x(t) = 18te^-t (the m meant its in meters not a variable right?)

Put e^-t into denominator and that makes it 18t/e^t

Using quotient rule:

f = 18t
f' = 18
g = e^t
g' = e^t

x'(t) = (f'g - fg')/g2
= (18e^t - 18te^t)/e^2t
factor out e^t and cancel it from the denom
= (18-18t)/e^t
 
  • #8
LowlyPion
Homework Helper
3,090
4
...don't know what m is... but I did try the quotient rule...and got different roots in the second term
m is meters.

But I think you are missing something.

N.B. (et)' = et
 
  • #9
27
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ok wow, this has officially turned into a calculus question, so forgive me.

First, I dont know how g(t) = g'(t).

Secondly, I think 18-18t = 0 is a critical number. Is e^t also a critial number?
 
  • #10
27
0
Finally, some guidance on the solution:

18-18t=0

t=1

X(1) = 18(1)/e1

= 18/2.7183

= 6.6218 m
 
  • #12
152
0
Is e^t also a critial number?

e is a positive number. A positive number to a power is still going to be always positive (ignoring limits at infinity). So you will (almost) always be able to cancel out exponents from an equation when it's equal to zero. You will see that done a lot if you take differential equations. Since it can't be equal to zero, it does not lead to a critical point as a solution.
 
  • #13
27
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not even as a vertical asymptote?
 
  • #14
152
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not even as a vertical asymptote?
That's why I covered my butt with "ignoring limits at infinity" :)

Are you sure you mean vertical? For critical critical points you want the derivative of the function = 0. A vertical asymptote would approach t=constant as the function derivative (velocity) went to +/- infinity. In any case, I don't believe the exponential has vertical asymptotes. It does have horizontal asymptotes (unless you are plotting t on the vertical axis and x on horizontal axis, in which case my horizontal becomes your vertical).

In your problem, the e^(-t) in the derivative goes to zero as t approaches infinity. Taking the limit of your original function as t goes to infinity shows that it also approaches zero.

This model then says, the electron starts out at x=0 with some positive velocity. As it proceeds in the positive x direction it is slowing down. It comes to a momentary stop at t=1, velocity=0. The velocity becomes negative, and the electron head back towards x=0. It is always slowing down as it heads towards x=0. It takes a infinite amount of time before it actually arrives back at exactly x=0 with velocity zero. Thus the problem was carefully worded with the phrase "momentarily at rest".

But, yes, you should be careful about what happens in the limits. That was excellent of you to consider it.
 
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