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d/dt (7s/2t3) = negative 21s/2t4 (ds/dt)
(Not bad for stuff I haven't used in 30 years, eh ?)
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∙ 11y ago
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If y equals x cube plus 2x and dx over dt equals 5 what is dy over dt when x equals 2?
y=x3+ 2x, dx/dt=5, x=2, dy/dt=? Differentiate the equation with respect to t. dy/dt=3x2*dx/dt Substitute in known values. dy/dt=3(2)2 * (5) dy/dt=60
How do you calculate net force equals speed equals velocity equals momentum equals acceleration equals?
Force (F) F = m.a and since a = dv/dt thus F = m.dv/dt Momentum (p) p = m.v and since a = dv/dt thus p = m.a.dt By switch dt from R.H.S. to L.H.S we get dp/dt = m.a thus F = dp/dt
Solve the differential equation dz over dt plus 4et plus z equals 0?
You need to clarify what you want to solve for. If you're solving for z, then we can say: dz/dt + 4et + z = 0 ∴ dz/dt = -4et - z ∴ ∫(dz/dt) dt = -2et2 - zt + C ∴ z = -2et2 - zt + C ∴ z + zt = -2et2 + C ∴ z(1 + t) = -2et2 + C ∴ z = (-2et2 + C) / (1 + t)
How is momentum and law of conservation of momentum related?
Momentum is the quantity that is conserved in this case. Conservation of Momentum is a consequence of Conservation of Energy, which equates to the sum of forces equals zero. 0 = f1 + f2 = dp1/dt + dp2/dt = d(p1 +p2)/dt = d(constant)/dt =0.
What is the condition for maximum velocity?
The condition for maximum velocity is acceleration equals zero; dv/dt = a= o.
Find the arc length L of the curve C given parametrically x equals 1 plus 2et y equals et 0 t1?
I will assume that you mean to ask, "What is the arc length of curve C from t=0 to t=1 if curve C is defined parametrically by x=1+2e^t and y=e^t?" I can answer this question. dx/dt=2e^t and dy/dt=e^t. Arc length = a∫b √[(dx/dt)2+(dy/dt)^2] = 0∫1 √[4e^(2t)+e^(2t)]dt = 0∫1 √[5e^(2t)]dt. = 0∫1 [(√5)(e^t)]dt = √5 x (e^1-e^0) = √5 x (e-1) = e√5-√5. Difficult? Maybe. Fun? Hopefully. Accurate? Definitely!
How do you diffentiate y equals 3.5 minus 3.5 times the Exp minus 1 over 20t?
y(t) = 3.5 - 3.5*e1-20t Then dy/dt = -3.5*e1-20t*(-20) = 70*e1-20t
How rockets engines work and relate on the third law of motion?
Rockets work on the conservation of vector energy, cP. 0 = dcP/dr = cdP/cdt=dP/dt = d(mV)/dt = mdV/dt + Vdm/dt=0 Thus, mdV/dt = -Vdm/dt, or (dV/dt)/V = -(dm/dt)/m. The Rocket's mass accelerates at the rate of the mass changes dm/dt.
What is derivative of cot t dt?
d/dt cot (t) dt = - cosec2(t)
How does Huygens find the acceleration in uniform circular motion?
a = dv/dt =d(vet)/dt =dv/dt *et+det/dt *vwith det =...
Can you have a zero velocity and zero displacement?
Yes, dD/dt = d0/dt = 0 thusDisplacement D=0 and Velocity dD/dt=d0/dt = o.
How do you build definition lists?
<dl> definition list <dt></dt> definition tag <dd></dd> data definition <dt></dt> <dd></dd> </dl>
