- #1
stunner5000pt
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Another long question but not that hard. Most of the writing is my work/questions
According to my prof if i cna solve this... the resulting relation can be used to solve Bernoulli's problem
For [tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
where [tex] \dot{y} = \frac{dy}{dx} [/tex]
Show that if [itex] F = F(y,\dot{y}) [/itex] only (not dependant of x explicitly) then the quantity
[tex] H = \dot{y} \frac{\partial F}{\partial \dot{y}} - F [/tex]
is a constant. That is [itex] \frac{dH}{dx} = 0 [/itex].
is this the total derivative (like it asks) of H wrt x? Since F does depend on y and y dot... which depend on x... then would hteir derivatives wrt x be included in the rightmost expression below?
[tex] \frac{dH}{dx} = \frac{d \dot{y}}{dx} \frac{\partial F}{\partial \dot{y}} + \dot{y} \frac{\partial F}{\partial \dot{y} dx} - \frac{dF}{dx} [/tex
is this true >> [tex] \frac{d \dot{y}}{\partial \dot{y}} =1 [/tex] ??
in that case all we're left with (if the dH/dx is correct) is
[tex] \frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx} [/tex]
if i were to differentiate the expression for F
[tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
wrt to y dot and x i would get zero, yes?
SO dH/dx = 0?
Determine the expression for H if F = 1/2 mv^2 - V(x) and explain the physical significance of this quantity.
is this correct? chain rule application everywhere
[tex] \frac{\partial F}{\partial \dot{y}} = m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} [/tex]
then [tex] H = \dot{y} m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} - \frac{1}{2} m \dot{x}^2 + V(x) [/tex]
the physical significance... hmm ...
H is a consnat quantity wrt x. But I am not sure how to interpret this past that part.
Please help! Thank you!
According to my prof if i cna solve this... the resulting relation can be used to solve Bernoulli's problem
For [tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
where [tex] \dot{y} = \frac{dy}{dx} [/tex]
Show that if [itex] F = F(y,\dot{y}) [/itex] only (not dependant of x explicitly) then the quantity
[tex] H = \dot{y} \frac{\partial F}{\partial \dot{y}} - F [/tex]
is a constant. That is [itex] \frac{dH}{dx} = 0 [/itex].
is this the total derivative (like it asks) of H wrt x? Since F does depend on y and y dot... which depend on x... then would hteir derivatives wrt x be included in the rightmost expression below?
[tex] \frac{dH}{dx} = \frac{d \dot{y}}{dx} \frac{\partial F}{\partial \dot{y}} + \dot{y} \frac{\partial F}{\partial \dot{y} dx} - \frac{dF}{dx} [/tex
is this true >> [tex] \frac{d \dot{y}}{\partial \dot{y}} =1 [/tex] ??
in that case all we're left with (if the dH/dx is correct) is
[tex] \frac{dH}{dx} = \dot{y} \frac{\partial F}{\partial \dot{y} dx} [/tex]
if i were to differentiate the expression for F
[tex] \delta \int_{x_{1}}^{x_{2}} F(x,y(x),\dot{y}(x)) dx = 0 [/tex]
wrt to y dot and x i would get zero, yes?
SO dH/dx = 0?
Determine the expression for H if F = 1/2 mv^2 - V(x) and explain the physical significance of this quantity.
is this correct? chain rule application everywhere
[tex] \frac{\partial F}{\partial \dot{y}} = m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} [/tex]
then [tex] H = \dot{y} m \dot{x} \frac{d \dot{x}}{d \dot{y}} - \frac{\partial V}{\partial \dot{y}} \frac{\partial x}{\partial \dot{y}} - \frac{1}{2} m \dot{x}^2 + V(x) [/tex]
the physical significance... hmm ...
H is a consnat quantity wrt x. But I am not sure how to interpret this past that part.
Please help! Thank you!