The velocity v of a freefalling skydiver is well modeled by the differential equation
m*dv/dt=mg-kv^2
where m is the mass of the skydiver, g is the gravitational constant, and k is the drag coefficient determined by the position of the driver during the dive.
(a)Find the general solution of the differential equation.
(b) Calculate the limit of v(t) as t -> INF.
I just started with dv/dt=g-(k/m)v^2 I don't really know where to go from there. Tried integrating dv/(g-(k/m)v^2) but I get stucked.
The answer for (a) is supposed to be v(t) = sqrt(m/(kv)*(Ce^(2*sqrt(kg/m)*t)-1)/(Ce^(2*sqrt(kg/m)*t)+1)
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and for (b)sqrt (mg/k)
you have dv/dt = F(v)
use separation of variables to get
v'/F(v) = 1
or
dv/F(v) = dx
Now, F(v) = g-(k/m)v^2 = k/m (a^2-v^2)
where a^2 = gm/k
That integrate into 1/a arctanh(v/a) = t
which is just a big exponential expression like what you show.
arctanh(z) = 1/2 ln((1+z)/(1-z))
The details get messy. If you go to
http://wood.mendelu.cz/math/maw-html/index.php?lang=en&form=ode
you can enter the equation (sorry, no letter constants, but you can use y'=1-2*y**2 to see the method). It will show you t as a function of v, with logs as above. Convert to v as a function of t to encounter the exponentials you want.
To find the general solution of the given differential equation, we need to separate the variables and solve the resulting first-order differential equation.
Starting with the equation you obtained: dv/dt = g - (k/m)v^2
Step 1: Separate the variables:
Rearrange the equation to separate variables on each side:
dv/(g - (k/m)v^2) = dt
Step 2: Integrate both sides:
Integrate with respect to v on the left side and with respect to t on the right side:
∫ 1/(g - (k/m)v^2) dv = ∫ dt
Step 3: Solve the integrals:
Now, let's solve each integral separately.
On the left side, we have:
∫ 1/(g - (k/m)v^2) dv
We can simplify this integral by using a trigonometric substitution. Let's substitute v = sqrt(mg/k) * tan(θ), which allows us to express the integral in terms of θ.
dv = sqrt(mg/k) * sec^2(θ) dθ
(g - (k/m)v^2) = (g - (k/m)(mg/k)tan^2(θ)) = g(1 - tan^2(θ)) = g * cos^2(θ)
Substituting these into the integral:
∫ 1/(g(1 - tan^2(θ))) * sqrt(mg/k) * sec^2(θ) dθ
=> √(m/(kg)) * ∫ sec^2(θ) dθ
=> √(m/(kg)) * tan(θ)
On the right side, we have:
∫ dt = t + C
Step 4: Combine the results:
Plugging back the substitutions, we have:
√(m/(kg)) * tan(θ) = t + C
Now, we need to solve for v in terms of t. Rearrange the equation to isolate tan(θ):
tan(θ) = (kg/m)^(1/2) * (t + C)
And since v = sqrt(mg/k) * tan(θ), we can substitute this expression into the equation above:
v = sqrt(mg/k) * (kg/m)^(1/2) * (t + C)
Simplifying, we get:
v = sqrt(m/(kg)) * (kg)^(1/2) * (t + C)
v = sqrt(m/(kg)) * sqrt(kg) * (t + C)
v = sqrt(m/k) * sqrt(g) * (t + C)
Let C' = sqrt(m/k) * C, we can simplify further:
v(t) = sqrt(m/k) * sqrt(g) * t + C'
So, the general solution for the differential equation is given by v(t) = sqrt(m/k) * sqrt(g) * t + C'.
For part (b), as t approaches infinity, the term sqrt(m/k) * sqrt(g) * t dominates the equation, and the constant term becomes negligible. Therefore, the limit of v(t) as t goes to infinity is sqrt(mg/k).
To solve the given differential equation, we need to separate variables and then integrate. Let's go through the steps:
(a) First, we begin with the differential equation:
m * dv/dt = mg - kv^2
Divide both sides by mg - kv^2:
dv / (mg - kv^2) = dt / m
Now, we'll integrate both sides:
∫ dv / (mg - kv^2) = ∫ dt / m
To integrate the left side, we can perform a trigonometric substitution. Let's substitute u = √(k * g / m) * v:
dv = (m / (k * g)) * du / u
Substituting back, we have:
∫ du / (u^2 - 1) = t / m
Now, we can integrate the left side by using the partial fraction decomposition:
∫ (1 / (u - 1) - 1 / (u + 1)) du = t / m
Now, integrating each term:
ln |(u - 1) / (u + 1)| = t / m + C
Taking the exponential of both sides:
|(u - 1) / (u + 1)| = e^(t / m + C)
Since e^C is just a constant, we can replace it with another constant, let's say A:
|(u - 1) / (u + 1)| = Ae^(t / m)
Now, we can split it into two cases:
Case 1: (u - 1) / (u + 1) = Ae^(t / m)
u - 1 = Au e^(t / m) + Ae^(t / m)
Case 2: (u - 1) / (u + 1) = -Ae^(t / m)
u - 1 = -Au e^(t / m) - Ae^(t / m)
Now, we solve each case separately:
Case 1: u - 1 = Au e^(t / m) + Ae^(t / m)
u = 1 + Ae^(t / m) + Au e^(t / m)
Substitute back u = √(k * g / m) * v:
√(k * g / m) * v = 1 + Ae^(t / m) + A√(k * g / m) * v * e^(t / m)
Rearrange the equation:
v - A√(k * g / m) * v * e^(t / m) = 1 + Ae^(t / m)
Factor out v:
v(1 - A√(k * g / m) * e^(t / m)) = 1 + Ae^(t / m)
Finally, we can isolate v by dividing both sides:
v(t) = (1 + Ae^(t / m)) / (1 - A√(k * g / m) * e^(t / m))
Simplifying the expression, we find the general solution for (a):
v(t) = √(m / (k * g)) * (1 + Ce^(2√(k * g / m) * t)) / (1 + Ce^(2√(k * g / m) * t))
where C = A^(2√(k * g / m))
(b) To calculate the limit of v(t) as t approaches infinity, we need to examine the exponential term in the numerator and denominator:
As t approaches infinity, e^(2√(k * g / m) * t) becomes larger and larger.
When e^(2√(k * g / m) * t) is very large, the terms with smaller magnitudes (1) become negligible.
Therefore, as t approaches infinity, the numerator becomes dominated by the exponential term, and the denominator also becomes dominated by the same exponential term.
Taking the limit as t goes to infinity, we can simplify the expression:
lim v(t) as t -> ∞ = √(m / (k * g)) * (1 + C∞) / (1 + C∞) = √(m / (k * g))
So, the limit of v(t) as t approaches infinity is √(m / (k * g)).