# Differentiating the equation for the mechanical energy of a spring

• ChiralSuperfields
In summary, the conversation discussed differentiating ##E = \frac {1}{2}mv^2 + \frac {1}{2}kx^2## with respect to time and the resulting answer being ##\frac {dE}{dt} = mva + kxv##. It was clarified that the correct derivative is ##\frac{dE}{dt}=mva+kxv##, and it was explained that this is due to the chain rule. The conversation also touched on the derivative of ##x^2## with respect to x, which is ##2x##, and how this relates to the chain rule in differentiating the original equation.
ChiralSuperfields
Homework Statement
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!

Callumnc1 said:
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##?

ChiralSuperfields
Chain rule. x and v are functions of t.

ChiralSuperfields
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.

Last edited:
ChiralSuperfields
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.

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.

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##

ChiralSuperfields
erobz said:
You almost certainly have, you just don’t realize it. ##y=x^2##

Oh I thought that was taking the derivative of y with respect to x to get ##2x##?

Callumnc1 said:

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##.

ChiralSuperfields
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} ##

erobz
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?

ChiralSuperfields
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##.

Last edited:
ChiralSuperfields
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:

$$\frac{dy}{dx}= 2x$$

ChiralSuperfields and haruspex
erobz said:
last line typo:

$$\frac{dy}{dx}= 2x$$
thanks - corrected,

ChiralSuperfields
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!

## 1. What is the equation for mechanical energy of a spring?

The equation for mechanical energy of a spring is E = 1/2kx^2, where E represents the mechanical energy, k is the spring constant, and x is the displacement from the equilibrium position.

## 2. How do you differentiate the equation for mechanical energy of a spring?

To differentiate the equation for mechanical energy of a spring, you must use the power rule of differentiation. This means you multiply the coefficient (1/2) by the exponent (2) and then subtract 1 from the exponent to get E' = kx.

## 3. What does the spring constant represent in the equation for mechanical energy of a spring?

The spring constant, k, represents the stiffness of the spring. It is a measure of how much force is needed to stretch or compress the spring by a certain distance.

## 4. How does displacement affect the mechanical energy of a spring?

The displacement, x, directly affects the mechanical energy of a spring. As the displacement increases, the mechanical energy also increases. This is because the more the spring is stretched or compressed, the more potential energy it has.

## 5. What is the significance of differentiating the equation for mechanical energy of a spring?

Differentiating the equation for mechanical energy of a spring allows us to calculate the rate of change of mechanical energy with respect to displacement. This can help us understand the behavior of the spring and how it responds to different forces and displacements.

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