Solving Inverse Functions in Electric Charge Equations

  • Thread starter Thread starter Null_
  • Start date Start date
  • Tags Tags
    Inverse
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 8K views
Null_
Messages
227
Reaction score
0

Homework Statement


1) When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by the following. (The maximum charge capacity is M and t is measured in seconds.)

Q(t) = M(1 - e-t/a)

(a) Find the inverse of this function and explain its meaning.


2. Let f(x) = 2 + x2 + tan(πx/2), where -1 < x < 1.
(a) Find f(f -1(4)).


Homework Equations



n/a

The Attempt at a Solution


(1) Q = M(1 - e^-t/a)
Q/M = 1 - e^-t/a
e^-t/a = 1 - Q/M
-t/a (ln e) = ln(1 - Q/M)
-t/a = ln(1 - Q/M)
t = -a[ln(1 - Q/M)]

That's not right.



(#2.) x = 2 + y^2 + tan(π*y/2)
4 = 2 + y^2 + tan(π*y/2)
*plug into wolfram alpha... y~0.642
f(x) = 2 + x2 + tan(πx/2) original formula
f(.642)= 2 + (.642^2) + tan (.642π/2)
f(.642)=2.4293


That's not right either.


I'm not really seeing where I went wrong in either of them..any help is appreciated.
 
Physics news on Phys.org
Null_ said:

Homework Statement


1) When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by the following. (The maximum charge capacity is M and t is measured in seconds.)

Q(t) = M(1 - e-t/a)

(a) Find the inverse of this function and explain its meaning.


2. Let f(x) = 2 + x2 + tan(πx/2), where -1 < x < 1.
(a) Find f(f -1(4)).


Homework Equations



n/a

The Attempt at a Solution


(1) Q = M(1 - e^-t/a)
Q/M = 1 - e^-t/a
e^-t/a = 1 - Q/M
-t/a (ln e) = ln(1 - Q/M)
-t/a = ln(1 - Q/M)
t = -a[ln(1 - Q/M)]
Why do you think this is wrong? This is what I got, but I am uncertain about the equation you wrote. I think you meant Q = M(1 - e^(-t/a)). If that is what you meant, then your answer is correct. However, you did not explain the meaning of the inverse here.
Null_ said:
That's not right.



(#2.) x = 2 + y^2 + tan(π*y/2)
4 = 2 + y^2 + tan(π*y/2)
*plug into wolfram alpha... y~0.642
f(x) = 2 + x2 + tan(πx/2) original formula
f(.642)= 2 + (.642^2) + tan (.642π/2)
f(.642)=2.4293


That's not right either.
Yes, you have an error here. You have the right idea, but probably made an error when you calculated f(.642). You should get a final answer that is close to 4. You'll get better results with more precision in your value for f-1(4), which I took as 0.64216.
Null_ said:
I'm not really seeing where I went wrong in either of them..any help is appreciated.
 
Mark44 said:
Why do you think this is wrong? This is what I got, but I am uncertain about the equation you wrote. I think you meant Q = M(1 - e^(-t/a)).
If that is what you meant, then your answer is correct. However, you did not explain the meaning of the inverse here.
I think it's wrong because it's an online homework assignment and it told me that my answer was wrong. :/ That is the right equation. Sorry, I got the explanation part right but just copied the whole question.
Mark44 said:
Yes, you have an error here. You have the right idea, but probably made an error when you calculated f(.642). You should get a final answer that is close to 4. You'll get better results with more precision in your value for f-1(4), which I took as 0.64216.
Ah, gotcha. Thanks.