QA: as x goes to infinity,
i) describe how 1-exp(-x) behaves
ii) describe how its derivative behaves
iii) describe how ln(x) behaves
iv) describe how the derivative of ln(x) behaves
v) Can we say "If the derivative goes to zero, then the original function stays bounded" as x goes to infinity? Explain.
QB: The Rocket Equation
The velocity of a rocket t seconds after liftoff from earth can be modeled by
v(t) = -g*t - v_e * Ln( (m-r*t)/m )
where
g = 9.8 m/s^2, the usual earth gravity value,
v_e = exhaust velocity, 3000 m/s (the e here isn't related to 2.71828...)
m = initial mass of the rocket+fuel: 30,000 kg
r = rate of using fuel = 160 kg/s
i) Find a formula for the acceleration function a(t) and graph it, t=0 to 60. Do the formula work by hand, but not the graphing, of course.
ii) Find a formula for the jerk function j(t) = a'(t) and graph it, t=0 to 60. Do the formula work by hand, but not the graphing.
iii) optional: what happens as t approaches 187.5 ? Explain. Hint: it doesn’t have anything to do with breaking free of gravity or of the atmosphere. Think about the “m” and “r” parameters
QA:
i) To describe how 1-exp(-x) behaves as x goes to infinity, we can analyze the limit of the function.
As x goes to infinity, exp(-x) approaches 0 since exponential functions decay very quickly. Therefore, 1-exp(-x) approaches 1 - 0 = 1. So, the function 1-exp(-x) approaches 1 as x goes to infinity.
ii) To describe how the derivative of 1-exp(-x) behaves as x goes to infinity, we need to find the derivative of 1-exp(-x) first.
Taking the derivative of 1-exp(-x) with respect to x gives us: d/dx (1-exp(-x)) = exp(-x).
As x goes to infinity, exp(-x) approaches 0, so the derivative of 1-exp(-x) also approaches 0. Therefore, the derivative of 1-exp(-x) approaches 0 as x goes to infinity.
iii) To describe how ln(x) behaves as x goes to infinity, we can analyze the limit of the function.
As x goes to infinity, ln(x) approaches infinity because logarithmic functions grow slowly as their input increases. So, ln(x) approaches infinity as x goes to infinity.
iv) To describe how the derivative of ln(x) behaves as x goes to infinity, we need to find the derivative of ln(x) first.
Taking the derivative of ln(x) with respect to x gives us: d/dx (ln(x)) = 1/x.
As x goes to infinity, 1/x approaches 0 since the denominator becomes very large. Therefore, the derivative of ln(x) approaches 0 as x goes to infinity.
v) We cannot say "If the derivative goes to zero, then the original function stays bounded" as x goes to infinity in general. This statement is not always true.
For example, consider the function f(x) = x^2. Its derivative, f'(x) = 2x. As x goes to infinity, f'(x) approaches infinity, but the original function f(x) = x^2 is unbounded (it keeps increasing without limit).
So, the behavior of the derivative alone does not guarantee that the original function will stay bounded.
QB:
i) To find the formula for the acceleration function a(t) and graph it, we need to differentiate the given velocity function, v(t), with respect to t.
The derivative of v(t) = -g*t - v_e * ln((m-r*t)/m) with respect to t is a(t) = -g - v_e * (r/m).
The formula for the acceleration function is a(t) = -g - v_e * (r/m).
ii) To find the formula for the jerk function j(t) = a'(t) and graph it, we need to differentiate the acceleration function, a(t), with respect to t.
The derivative of a(t) = -g - v_e * (r/m) with respect to t is j(t) = 0.
The formula for the jerk function is j(t) = 0, which means that the jerk remains constant at zero.
iii) As t approaches 187.5, the rocket will have consumed all its fuel, resulting in the mass becoming zero. In the rocket equation, when the mass (m) becomes zero, the logarithmic term becomes undefined.
Therefore, as t approaches 187.5, the rocket equation is no longer valid since the logarithmic term is undefined, and the model breaks down. This has nothing to do with breaking free of gravity or the atmosphere, but rather the limitation of the model itself when the mass becomes zero.
These look like pretty standard problems. What is giving you trouble? Which parts have you tried, and with what results
You know what the graphs of e^-x and ln(x) look like, so start from there.
acceleration is just dv/dt, so start there.