Use chain rule:Solve dF/dt given: F(x,y)=-6x^3-8y^5
where x(t)=-4t^4 and y(t)=6t
1st use equal signs instead of 'is'So far I only have achieve getting to dF/dx[-4t^4] is -16t
dF/dy[6t] is 6
And I can't figure out how to do dx/dt or dy/dt
Ans i need help with how to put it into the chain rule to so
Would I just take a second derivative of 4t^4 to get dx/dt?If x=-4t^4, why is the trouble with finding dx/dt
Could you find dy/dx for y=-4x^4?
I think you're so lost. Ignore t for now. Compute:Would I just take a second derivative of 4t^4 to get dx/dt?
So, dF/dx[4t^4]=16t^3
dx/dt[16t^3]=48t^2
?
I am indeed so lost. SoI think you're so lost. Ignore t for now. Compute:
[math]\frac{\partial F}{\partial x}=\frac{\partial(6x^3-8y^5)}{\partial x}=?\\ \frac{\partial F}{\partial y}=\frac{\partial(6x^3-8y^5)}{\partial y}=?\\[/math]
So...If you are not asked to use the chain rule, you can do it directly by substituting the values of \(\displaystyle x\) and \(\displaystyle y\) in \(\displaystyle F\).
\(\displaystyle F(x,y)\) will become \(\displaystyle F(t)\).
\(\displaystyle F(t) = -6(-4t^4)^3 - 8(6t)^5 = 384t^{12} - 62208t^5\)
Now you can find \(\displaystyle \frac{dF}{dt}\) in one shot.
But if I were you, I would always use the chain rule. It is more fancy and it allows you to solve complicated functions without any headaches.
\(\displaystyle \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}\)
If this partial symbol \(\displaystyle \partial\) makes you confused, you can remove it and assume the derivative like this,
\(\displaystyle \frac{dF}{dt} = \frac{d F}{d x}\frac{dx}{dt} + \frac{d F}{d y}\frac{dy}{dt}\)
The brain is supposed to function better now
I don't know how to do this...would I take the derivative of each one so the first answer would be: 18x^2-8y^5?I think you're so lost. Ignore t for now. Compute:
[math]\frac{\partial F}{\partial x}=\frac{\partial(6x^3-8y^5)}{\partial x}=?\\ \frac{\partial F}{\partial y}=\frac{\partial(6x^3-8y^5)}{\partial y}=?\\[/math]
When you are computing partial derivative respect to x, treaty y as a constant and vice versa:I don't know how to do this...would I take the derivative of each one so the first answer would be: 18x^2-8y^5?
And the second would be 6x^3-40y^5?
You missed the negative. Understanding what derivatives mean will help you be less confused.So for the second part dx/dt=-16t^3
And dy/dt=6?
Look at the original post you made, what does x equal and what does y equal?OK great thank you! Would this be correct?
dF/dt=(18x^2)*(-16t^3)+(-40y^4)(6)
And you cannot combine them can you?
If I am correct, you will have to buy me a bar of chocolate?I am indeed so lost. So
So...
dF/dt= 4608t^11-311,040t^4??
This just seems wrong
F = 6*x^3 - 8*y^5Use chain rule:
dF/dt = dF/dx * dx/dt + dF/dy * dy/dt
So:where x(t)=-4t^4 and y(t)=6t
But it's x(t)...and y(t) so I would want to plug in those equations for t?Look at the original post you made, what does x equal and what does y equal?
Correct. The goal is to put everything in terms of t, hence the meaning of the notation [imath]\frac{dF}{\red{dt}}[/imath], i.e. how is F changed as t changes?But it's x(t)...and y(t) so I would want to plug in those equations for t?