The potential energy function of a particle moving in one-dimension is

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The potential energy function is given as U = k(x^2 + y^2), and the force exerted on the particle can be determined using the formula F = -∇U. The user attempts to calculate the force but expresses uncertainty about correctly applying partial derivatives. It is clarified that when taking partial derivatives, other variables must be treated as constants, leading to the correct derivatives of -2kx for x and -2ky for y. The final expression for the force is F = -2kx ihat - 2ky jhat.
pittuniv
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Homework Statement



U = k(x^2 + y^2) What is the force exerted on the particle?

Homework Equations



F = -(¶U/¶x ihat +¶U/¶y jhat +¶U/¶z khat) <--couldnt get the del symbol right
determining force from potential energy

The Attempt at a Solution



F = -¶/¶x[k(x^2+y^2)]ihat - ¶/¶y[k(x^2 +y^2)]jhat
=-[2kx + Y^2]ihat - [kx^2 + 2ky]jhat

Im new to calculus, and I am pretty sure that I am not doing the derivative of this right...any help would be greatly appreciated.
 
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When you're doing partial derivatives of a function of several variables (x,y,z,...) with respect to a given variable, all the other independent variables are treated as constants. So,

-\frac{\partial U}{\partial x} = -2 k x ~~~~~~~~~~-\frac{\partial U}{\partial y} = -2 k y
 
pittuniv said:

Homework Statement



U = k(x^2 + y^2) What is the force exerted on the particle?

Homework Equations



F = -(∂U/∂x ihat +∂U/∂y jhat +∂U/∂z khat) <--couldnt get the del symbol right
determining force from potential energy

The Attempt at a Solution



F = -∂/∂x[k(x^2+y^2)]ihat - ∂/∂y[k(x^2 +y^2)]jhat
=-[2kx + Y^2]ihat - [kx^2 + 2ky]jhat

Im new to calculus, and I am pretty sure that I am not doing the derivative of this right...any help would be greatly appreciated.
The partial derivative with respect to one variable is taken with the other variable held constant. So \partial U/\partial x = 2kx\hat x and \partial U/\partial y = 2ky\hat y.

AM
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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