What is the derivative of ##x^2## wrt ##x##?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!
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?haruspex said:What is the derivative of ##x^2## wrt ##x##?
Thank you for your reply @Frabjous !Frabjous said:Chain rule. x and v are functions of t.
Thank you for your reply @erobz !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.
You almost certainly have, you just don’t realize it. ##y=x^2##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?
Thanks for your reply @erobz!erobz said:You almost certainly have, you just don’t realize it. ##y=x^2##
Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##erobz said:That’s correct. Then you apply the chain rule. First differentiate ##y =x^2 ## wrt ##x##, then ##x## wrt ##t##.
Not quite. You must do the same to each side of an equation.Callumnc1 said:Thank you for your reply @erobz! I think it would be ## y = (2x)\frac {dx}{dt} ##
last line typo: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##.
thanks - corrected,erobz said:last line typo:
$$ \frac{dy}{dx}= 2x $$
Thank you for your replies @erobz and haruspex! Sorry, that was a silly mistake I should not have made!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##.
The mechanical energy of a spring is given by the equation \( E = \frac{1}{2} k x^2 \), where \( E \) is the mechanical energy, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
To differentiate the mechanical energy equation \( E = \frac{1}{2} k x^2 \) with respect to displacement \( x \), you apply the power rule of differentiation. The derivative is \( \frac{dE}{dx} = kx \).
The derivative \( \frac{dE}{dx} = kx \) represents the force exerted by the spring. This is consistent with Hooke's Law, which states that the force exerted by a spring is proportional to its displacement, \( F = -kx \). The negative sign indicates that the force is restorative, acting in the direction opposite to the displacement.
The spring constant \( k \) is a measure of the stiffness of the spring. It remains constant during differentiation and directly affects the magnitude of the force calculated from the derivative. A higher \( k \) means a stiffer spring, which produces a larger force for a given displacement.
For springs with non-linear characteristics, the mechanical energy equation may not be \( E = \frac{1}{2} k x^2 \). Instead, a different potential energy function that accurately describes the spring's behavior must be used. The differentiation process would then be applied to this new function to find the corresponding force. Non-linear springs require a specific form of the potential energy function based on their unique properties.