Differentiating the equation for the mechanical energy of a spring

[Callumnc1]

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!

 

[haruspex]

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 $x$?

 

[Frabjous]

Chain rule. x and v are functions of t.

 

[erobz]

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.

 

[Callumnc1]

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?

 

[Callumnc1]

Chain rule. x and v are functions of t.

Thank you for your reply @Frabjous !

 

[Callumnc1]

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 !

 

[erobz]

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. $y=x^2$

 

[Callumnc1]

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 $2x$?

 

[erobz]

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 $t$.

 

[Callumnc1]

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]

Thank you for your reply @erobz! I think it would be $y = (2x)\frac {dx}{dt}$

Do you see how it works out?

 

[haruspex]

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

 

[erobz]

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:

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

 

[haruspex]

last line typo:

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

thanks - corrected,

 

[Callumnc1]

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!