Differential
- Thread starter Viona
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\(\displaystyle \dfrac{m v_{z_0}^2(t)}{2} = \dfrac{m v_z^2(t)}{2} + m g z\)
differentiate both sides with respect to \(\displaystyle t\)
\(\displaystyle m v_{z_0}(t) \dfrac{d v_{z_0}}{dt} = m v_z(t) \dfrac{d v_z}{dt} + 0\)
divide by \(\displaystyle m\), drop the \(\displaystyle t\)'s, and use the abbreviated form of the derivative since both sides are with respect to \(\displaystyle t\)
\(\displaystyle v_{z_0} dv_{z_0} = v_z d_{v_z}\)
mv2z0(t)=mv2z(t)+mgz
v2z0(t)=v2z(t)+gz
2vz0dvz0(t)=2vzdvz(t)+0 ----integration step, before this step and after is algebra----
2vz0dvz0(t)=2vzdvz(t)
vz0dvz0=vzdvz
Steven G
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No special rules at all. First divide the equation by m. Then use the use the basic power rule ( (xn)' = nxn-1). Lastly note the mgz is a constant so it's derivative is 0.
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Viona
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Thank you all for your help.
But I need to ask does t is short for time?
does this mean the variables I used is a function of time?
Another question: When we integrate a function of x with respect to x sometimes we need to change the variable to y (for instance: to simplify the integration) where y =ax and a is a constant. we say that dy=a dx. Is it the same procedure you all used above to answer my first question?
But I need to ask does t is short for time?
does this mean the variables I used is a function of time?
Another question: When we integrate a function of x with respect to x sometimes we need to change the variable to y (for instance: to simplify the integration) where y =ax and a is a constant. we say that dy=a dx. Is it the same procedure you all used above to answer my first question?
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Asking for "Fisty" fight? with Denis?