Instantaneous velocity electron

[jpd5184]

Homework Statement

an electron moving along the x-axis has a position given by x=19te^-4t m, where t is in seconds. how far is the electron from the origin when it momentarily stops.

Homework Equations

velocty=delta x / delta t

The Attempt at a Solution

i think its a instantaneous velocity question but I am not sure. i think you have to take the derivative of the equation. If somebody could just head me in the right direction then i could attempt to solve it.

 

[Doc Al]

i think its a instantaneous velocity question but I am not sure. i think you have to take the derivative of the equation.

So far, so good. Keep going.

 

[jpd5184]

so would the derivative of by x=19te^-4t m be:

-4(19te)^-3 which would be 76te^-3.

so i think i know what I am doing this far but I am not sure what the question is asking when it says
"how far is the electron from the origin when it momentarily stops."

how do i know the time when it momentarily stops. i mean if i know the time i can just substitute the time in for t and solve the equation to get the answer.

 

[Doc Al]

so would the derivative of by x=19te^-4t m be:

-4(19te)^-3 which would be 76te^-3.

Not exactly. Hint: You'll need the product rule. (I assume the expression is 19t e^-4t.)

how do i know the time when it momentarily stops. i mean if i know the time i can just substitute the time in for t and solve the equation to get the answer.

Hint: What's the speed when it 'momentarily stops'? Use that to solve for the time.

 

[jpd5184]

Hint: What's the speed when it 'momentarily stops'? Use that to solve for the time.[/QUOTE]

im guessing it would be zero.and there is no gap in between the 19t and the e^-4t. its just one whole expression.

 

[Doc Al]

im guessing it would be zero.

Exactly.

and there is no gap in between the 19t and the e^-4t. its just one whole expression.

Right. I wrote it with a gap for clarity, so the t and e wouldn't blur together. Redo your derivative.

 

[iamalexalright]

Doc Al included the gap to make it clear that you need to use the product rule

 

[jpd5184]

using the product rule i got.

f(x)= 19t
g(x)= e^-4t

so would it be (im not sure of the derivative of e^-4t)

19e^-4t + 19te^-4t

 

[iamalexalright]

Have to use the chain rule with e^(-4t)

Your answer is incorrect but close

 

[jpd5184]

using the chain rule i got the derivative of e^-4t to be ln(t)t^4

f(y)=e^4
g(x)=-4ln(t)

(ln(t))e^-4ln(t)
(ln(t))t^4

 

[gerben]

To use the chain rule you need to view the function, g(t) = e^-4t, in two parts like this:
g(u) = e^u, with u(t)=-4t

then using the chain rule to to get the derivative of g(t), dg/dt, means using this rule:
dg/dt = dg/du * du/dt

 

[jpd5184]

sorry but I am not really great at calculus, since I am biology major.

don't really know what u mean or what to do to use the chain rule.

 

[gerben]

well let's do it step by step, what is the derivative of:
e^u

 

[jpd5184]

not sure what u is but:

ln u

 

[gerben]

I just used 'u' here instead of the usual 'x'. It was meant to be the same question as "what is the derivative of e^x".

The derivative of a function is: the change of the value of the function per change in its variable.

So the "derivative of e^u", means the change of e^u per change in u.

For simple functions like f(x)=3x, it is clear that the change of f(x) per change in x is 3. since whenever x change by 1, f(x) changes by 3.

The notation I used above for the derivative of f(x) was df/dx, which you could read as "the change of f(x) per change in x", which is also often written as f'(x).

Anyway, for some functions the derivative is not immediately clear and you may just need to know what it is. The function f(x) = e^x, is a very special function because its derivative, df/dx or f'(x) is simply equal to the function itself.

 

[jpd5184]

To use the chain rule you need to view the function, g(t) = e^-4t, in two parts like this:
g(u) = e^u, with u(t)=-4t

then using the chain rule to to get the derivative of g(t), dg/dt, means using this rule:
dg/dt = dg/du * du/dt

so the derivative of g(u)=e^u is just e

and the derivative of u(t)=-4t is -4

 

[gerben]

To use the chain rule you need to view the function, g(t) = e^-4t, in two parts like this:
g(u) = e^u, with u(t)=-4t

so using the chainrule here you would get the derivative of g(t) as
g'(u) * u'(t)

I already told you that g'(u)=g(u) so that is just e^u=e^-4t.

now you just need to multiply this by the derivative of u(t)=-4t

 

[gerben]

so the derivative of g(u)=e^u is just e

and the derivative of u(t)=-4t is -4

the derivative of g(u)=e^u is e^u, the derivative of u(t) is correct

 

[jpd5184]

Homework Statement

an electron moving along the x-axis has a position given by x=19te^-4t m, where t is in seconds. how far is the electron from the origin when it momentarily stops.

The particle stops when the velocity is zero. The velocity function is the time derivative of the position function.

the derivative of the time function is (e^-4t)(-4t). so all i did was plug 0 in for t and i got my answer to be 0. but that's wrong, why?

 

[gerben]

The derivative of the position function is not correct, you were going in the right direction here in an earlier post:

using the product rule i got.

f(x)= 19t
g(x)= e^-4t

so would it be (im not sure of the derivative of e^-4t)

19e^-4t + 19te^-4t

It is just the second term that is not correct here, it is not 19te^-4t, it should be the 19t times the derivative of e^-4t

 

[jpd5184]

so it would be:

19e^-4t + 19te

which would be:

19e^-4(0) + 19te =

19 + 19(0)e = 19 + 0 = 0

 

[Doc Al]

so it would be:

19e^-4t + 19te

No.

x = (19t)(e^-4t)

What's the derivative of (19t)?

What's the derivative of (e^-4t)? (You need the chain rule here.)

Combine these using the product rule.

 

[jpd5184]

(-4 log e) / e^4t

then would i plug 0 in for t

 

[Doc Al]

(-4 log e) / e^4t

then would i plug 0 in for t

Answer the questions I asked in my last post.

 

[jpd5184]

No.

x = (19t)(e^-4t)

What's the derivative of (19t)?

What's the derivative of (e^-4t)? (You need the chain rule here.)

Combine these using the product rule.

the derivative of 19t is 19 but I am not sure the derivative of e^-4t, sorry but I am struggling.

 

[Doc Al]

the derivative of 19t is 19 but I am not sure the derivative of e^-4t, sorry but I am struggling.

OK.

Let f = e^g
Let g = -4t

df/dg = e^g (This is a key fact about the exponential function: its derivative equals itself. Memorize it!)

What's dg/dt?

Using the chain rule, what's df/dt? (Which will be the derivative of e^-4t.)

Hint: df/dt = (df/dg)(dg/dt)

 

[jpd5184]

thanks for everybody trying to help me but i really don't understand the derivation process.