Differentiating the equation for the mechanical energy of a spring

AI Thread Summary
Differentiating the mechanical energy equation, E = (1/2)mv² + (1/2)kx², with respect to time yields dE/dt = mva + kxv, not dE/dt = ma + kv. The confusion arises from applying the chain rule correctly, as both x and v are functions of time. The right-hand side of the incorrect equation is dimensionally inconsistent, combining forces inappropriately. Understanding the derivative of x² with respect to x is essential, as it leads to the correct application of the chain rule. The discussion emphasizes the importance of proper differentiation techniques in physics.
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Homework Statement
Please see below
Relevant Equations
## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ##
Why when we differentiate ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ## with respect to time the answer is ## \frac {dE}{dt} = mva + kxv ##?

I though it would be ##\frac {dE}{dt} = ma + kv ##.

Many thanks!
 
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Callumnc1 said:
Homework Statement:: Please see below
Relevant Equations:: ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ##

Why when we differentiate ## E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2 ## with respect to time the answer is ## \frac {dE}{dt} = mva + kxv ##?

I though it would be ##\frac {dE}{dt} = ma + kv ##.

Many thanks!
What is the derivative of ##x^2## wrt ##x##?
 
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Chain rule. x and v are functions of t.
 
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The time rate of change in energy is power.

With what you thought it would be the units on the RHS are Force added to Force per unit time. Not only are neither of them power, they are also dimensionally inconsistent with each other.
 
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haruspex said:
What is the derivative of ##x^2## wrt ##x##?
Thank you for your reply @haruspex ! I don't think I've ever taken the derivative of ##x^2## wrt ##x##. I think I've only the derivative of ##y## wrt ##x##. How would I take the derivative?
 
Frabjous said:
Chain rule. x and v are functions of t.
Thank you for your reply @Frabjous !
 
erobz said:
The time rate of change in energy is power.

With what you thought it would be the units on the RHS are Force added to Force per unit time. Not only are neither of them power, they are also dimensionally inconsistent with each other.
Thank you for your reply @erobz !
 
Callumnc1 said:
Thank you for your reply @haruspex ! I don't think I've ever taken the derivative of ##x^2## wrt ##x##. I think I've only the derivative of ##y## wrt ##x##. How would I take the derivative?
You almost certainly have, you just don’t realize it. ##y=x^2##
 
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erobz said:
You almost certainly have, you just don’t realize it. ##y=x^2##
Thanks for your reply @erobz!

Oh I thought that was taking the derivative of y with respect to x to get ##2x##?
 
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Callumnc1 said:
Thanks for your reply @erobz!

Oh I thought that was taking the derivative of y with respect to x to get ##2x##?
That’s correct. Then you apply the chain rule. First differentiate ##y =x^2 ## wrt ##x##, then ##x## wrt ##t##.
 
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  • #11
erobz said:
That’s correct. Then you apply the chain rule. First differentiate ##y =x^2 ## wrt ##x##, then ##x## wrt ##t##.
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
 
  • #12
Callumnc1 said:
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
Do you see how it works out?
 
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  • #13
Callumnc1 said:
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=2x##.
 
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  • #14
haruspex said:
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=x^2##.
last line typo:

$$ \frac{dy}{dx}= 2x $$
 
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  • #15
erobz said:
last line typo:

$$ \frac{dy}{dx}= 2x $$
thanks - corrected,
 
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  • #16
haruspex said:
Not quite. You must do the same to each side of an equation.
The derivative of y wrt x is ##\frac{dy}{dx}##.
The derivative of ##x^2## wrt x is ##\frac{d(x^2)}{dx}=2x##.
So differentiating both sides of ##y=x^2## wrt x gives
##\frac{dy}{dx}=2x##.
Thank you for your replies @erobz and haruspex! Sorry, that was a silly mistake I should not have made!
 
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