The rate of change in the volume of a tank is known to be dV/dt=0.6t*cos(0.08t^2 -1), where V(t) is in gallons per minute and 0<=t<=10 (0 is less than or equal to t which is less than or equal to 10). If the tank has a volume of 14 gallons initially, what is its volume at 10 minutes?
dv/dt = 0.6t cos(0.08t^2-1)
v = 3/8 sin(0.08t^2-1) + c
Since v(0) = 14, c = 14+3/8 sin(1) = 14.3
v(t) = .375 sin(.08t^2-1)+14.3
Now just evaluate v(10)
To find the volume of the tank at 10 minutes, we need to integrate the given rate of change expression with respect to time from 0 to 10.
∫(dV/dt) dt = ∫(0.6t*cos(0.08t^2 - 1)) dt
To compute this integral, we can use integration by substitution.
Let u = 0.08t^2 - 1
Therefore, du/dt = 0.16t
Rearranging, we get dt = du / (0.16t)
Substituting this value and the expression for du in the integral, we have:
∫(0.6t*cos(0.08t^2 - 1)) dt = ∫(0.6*cos(u)) * (du / (0.16t))
Simplifying, we get:
= (0.6/0.16) * ∫(cos(u)) * (du / t)
= 3.75 * ∫(cos(u)) * (du / t)
Now, integrating cos(u) with respect to u:
= 3.75 * sin(u)
Substituting back for u:
= 3.75 * sin(0.08t^2 - 1)
To find the volume at 10 minutes, we evaluate this expression at t = 10:
Volume at t=10 = 3.75 * sin(0.08(10)^2 - 1)
= 3.75 * sin(0.8 - 1)
= 3.75 * sin(-0.2)
= 3.75 * (-0.19866933) (rounded to 8 decimal places)
So, the volume of the tank at 10 minutes is approximately -0.745 (rounded to 3 decimal places).
To find the volume of the tank at 10 minutes, we can use the given rate of change equation and integrate it with respect to time. The integration will give us the volume function V(t), which we can then evaluate at t = 10 to find the volume at that time.
Given: dV/dt = 0.6t * cos(0.08t^2 - 1)
To find V(t), integrate both sides of the equation with respect to t:
∫ dV/dt dt = ∫ 0.6t * cos(0.08t^2 - 1) dt
Integrating the left-hand side is straightforward:
∫ dV/dt dt = ∫ dV = V(t)
For the right-hand side, we have the product of two functions: 0.6t and cos(0.08t^2 - 1). To integrate this, we need to use a technique called integration by substitution.
Let u = 0.08t^2 - 1
Differentiating u with respect to t: du/dt = 0.16t
Rearranging the equation, we have: dt = du / (0.16t)
Substituting the new variable and adjusting the limits of integration:
∫ 0.6t * cos(0.08t^2 - 1) dt = ∫ 0.6t * cos(u) * (du / (0.16t))
Simplifying: ∫ 0.6 * cos(u) * (du / 0.16)
Canceling out the t's and simplifying further: ∫ 3.75 * cos(u) du
Integrating the right-hand side:
∫ 3.75 * cos(u) du = 3.75 * sin(u) + C
Now, substituting back the original variable:
V(t) = 3.75 * sin(0.08t^2 - 1) + C
We can solve for C using the initial volume condition where t = 0 and V(t) = 14 gallons:
14 = 3.75 * sin(0.08(0)^2 - 1) + C
Simplifying: 14 = 3.75 * sin(-1) + C
Using the sine of -1: 14 = 3.75 * (-0.841) + C
Solving for C: C = 14 + 3.164
Now, we can rewrite V(t) with the determined C value:
V(t) = 3.75 * sin(0.08t^2 - 1) + 17.164
To find the volume at t = 10 minutes, substitute t = 10 into V(t):
V(10) = 3.75 * sin(0.08(10)^2 - 1) + 17.164
Calculating the right-hand side will give you the final answer, which represents the volume of the tank at 10 minutes.