Problem 25 - Advanced Engineering Mathematics

Question

While the object is in the air its velocity is modelled by the linear differential equation \(m d v / d t=m g-k v .\) Using \(m=160, k=\frac{1}{4},\) and \(g=32,\) the differential equation becomes \(d v / d t+(1 / 640) v=32 .\) The integrating factor is \(e^{\int d t / 640}=e^{t / 640}\) and the solution of the differential equation is \(e^{t / 640} v=\int 32 e^{t / 640} d t=20,480 e^{t / 640}+c\) Using \(v(0)=0\) we see that \(c=-20,480\) and \(v(t)=20,480-20,480 e^{-t / 640} .\) Integrating we get \(s(t)=20,480 t+\) \(13,107,200 e^{-t / 640}+c .\) since \(s(0)=0, c=-13,107,200\) and \(s(t)=-13,107,200+20,480 t+13,107,200 e^{-t / 640}\) To find when the object hits the liquid we solve \(s(t)=500-75=425,\) obtaining \(t_{a}=5.16018 .\) The velocity at the time of impact with the liquid is \(v_{a}=v\left(t_{a}\right)=164.482 .\) When the object is in the liquid its velocity is modeled by the nonlinear differential equation \(m d v / d t=m g-k v^{2} .\) Using \(m=160, g=32,\) and \(k=0.1\) this becomes \(d v / d t=\left(51,200-v^{2}\right) / 1600 .\) Separating variables and integrating we have \\[ \frac{d v}{51,200-v^{2}}=\frac{d t}{1600} \quad \text { and } \quad \frac{\sqrt{2}}{640} \ln \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=\frac{1}{1600} t+c \\] Solving \(v(0)=v_{a}=164.482\) we obtain \(c=-0.00407537 .\) Then, for \(v<160 \sqrt{2}=226.274\) \\[ \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=e^{\sqrt{2} t / 5-1.8443} \quad \text { or } \quad-\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}=e^{\sqrt{2} t / 5-1.8443} \\] Solving for \(v\) we get \\[ v(t)=\frac{13964.6-2208.29 e^{\sqrt{2} t / 5}}{61.7153+9.75937 e^{\sqrt{2} t / 5}} \\] Integrating we find \\[ s(t)=226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right)+c \\] Solving \(s(0)=0\) we see that \(c=3185.78,\) so \\[ s(t)=3185.78+226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right) \\] To find when the object hits the bottom of the tank we solve \(s(t)=75,\) obtaining \(t_{b}=0.466273 .\) The time from when the object is dropped from the helicopter to when it hits the bottom of the tank is \(t_{a}+t_{b}=\) 5.62708 seconds.

Step-by-Step Solution

Answer

The total time until the object hits the tank bottom is 5.62708 seconds.

Step 1: Understanding the Problem

We have a two-part problem. First, an object falls in the air modeled by a linear differential equation, and then it enters a liquid modeled by a nonlinear differential equation. We need to determine the total time taken from the drop until it hits the bottom of a tank.

Step 2: Modeling in the Air

We start with the differential equation in the air: \( \frac{dv}{dt} + \frac{1}{640}v = 32 \). The integrating factor is \( e^{t/640} \), leading to the solution \( v(t) = 20480 - 20480 e^{-t/640} \). We integrate again to find the position: \( s(t) = -13107200 + 20480t + 13107200e^{-t/640} \).

Step 3: Find Time to Reach Liquid (\(t_a\))

Set \( s(t) = 425 \) (accounting for initial position of 75 units): \( 20480t + 13107200e^{-t/640} = 13107625 \). By solving, we find \( t_a = 5.16018 \) seconds.

Step 4: Modeling in the Liquid

In the liquid, the equation is \( \frac{dv}{dt} = \frac{51200 - v^2}{1600} \). Separating variables and integrating results in the implicit solution and \( c = -0.00407537 \), obtained using initial condition \( v(0) = 164.482 \).

Step 5: Solve for Velocity in Liquid

We solve \( v(t) = \frac{13964.6 - 2208.29 e^{\sqrt{2} t / 5}}{61.7153 + 9.75937 e^{\sqrt{2} t / 5}} \) after manipulating the implicit equation and given conditions in the liquid.

Step 6: Integrate for Position in Liquid

Integrate the expression for \(v(t)\) to find position \(s(t) = 3185.78 + 226.275t - 1600\ln(6.3237 + e^{\sqrt{2}t/5})\). Solve for \(c\) using \(s(0) = 0\).

Step 7: Find Time to Reach Bottom (\(t_b\))

Set \( s(t) = 75 \), solve for \( t_b \), find \( t_b = 0.466273 \) seconds.

Step 8: Total Time Calculation

Add \( t_a \) and \( t_b \) to find total time: \( t_{total} = t_a + t_b = 5.16018 + 0.466273 = 5.62708 \) seconds.

Key Concepts

Linear Differential Equations

A linear differential equation involves an unknown function and its derivatives. In the problem, we initially deal with one linear differential equation in the air before transitioning into nonlinear terms when the object enters the liquid. The general form is given by the equation \( \frac{dv}{dt} + \frac{1}{640}v = 32 \). Here, \( v \) is the velocity and \( t \) is the time. The equation describes how the velocity of an object changes over time. Linear equations like these assume the influence on the velocity is directly proportional to the velocity itself, and any nonlinearity is absent.

Solving such equations often requires finding an integrating factor—which in this case is \( e^{t/640} \). This factor simplifies the process of integrating and solving the differential equation to find our function \( v(t) \). Understanding this approach is crucial when applying linear differential equations to real-world problems such as modeling velocity.

Nonlinear Differential Equations

Nonlinear differential equations are more complex as they allow for a relationship that is not directly proportional. In our exercise, the object entering a liquid is modeled using such an equation: \( \frac{dv}{dt} = \frac{51200 - v^2}{1600} \). This is significantly different from the linear case, as now the square of the velocity, \( v^2 \), plays a role in how velocity changes over time.

Such nonlinearity requires different methods to solve, often involving separation of variables and integration which can be a bit more challenging. These equations are often used in modeling scenarios where physical systems exhibit nonlinearity, capturing dynamics more accurately than linear models. Understanding and solving nonlinear differential equations is necessary for studying complex motion and behavior in various scientific fields.

Integration in Calculus

Integration is a fundamental concept in calculus used to find a function from its derivative or to calculate areas under curves. In the context of differential equations, integration is employed to solve equations that describe dynamic changes. Whether it is recovering velocity from acceleration or position from velocity, integration is the tool for obtaining these solutions.

In our problem, we start with a velocity model and integrate to determine the position over time. The integral \( \int 32 e^{t/640} dt = 20480 e^{t/640} + c \) allowed us to find the function \( v(t) \). Similarly, integrating further helps us go from velocity to position \( s(t) \), revealing how an object's position changes as its velocity evolves. Mastery of integration is essential to solve differential equations and thus study motion comprehensively.

Modeling Motion

Modeling motion through differential equations is a practical application of calculus. By using equations to describe how an object's velocity and position change over time, we can predict its future state under various conditions. This has broad applications in physics and engineering.

In our exercise, initially, we have an object falling through the air where resistance is proportional to velocity—captured through a linear equation. Upon entering the liquid, nonlinearity is introduced, reflecting the more complex interactions within the medium.

• For each phase—air and liquid—the model differs to capture the true impact of the medium on motion.
• Predictive models like this allow scientists and engineers to simulate and analyze scenarios without direct experimentation.

To effectively model motion, one must understand both the theoretical framework and computation of differential equations used in various contexts.

Separation of Variables

The separation of variables is a common technique for solving differential equations, particularly useful for equations where the variables can be isolated on opposite sides of the equation. It involves rearranging an equation so that all terms involving one variable appear on one side of the equation, and all terms of the other variable appear on the opposite side.

In our specific exercise, we separated variables in the nonlinear case when the object was in the liquid: \[ \frac{dv}{51200 - v^2} = \frac{dt}{1600} \]. We then integrated both sides to move closer to a solution. This allowed us to obtain an expression connecting \( v \) and \( t \), ultimately leading to a complete solution for the velocity in the liquid.

• Separation of variables is particularly valuable because it often simplifies complex problems into manageable integrals.
• Understanding this technique opens the door to handle a wide array of differential equations typically encountered in physics and engineering.

With practice, one becomes adept at recognizing when and how to employ this method effectively.