Three Questions In The Study of Calculus..?

Question

Anyone who can answer all three questions first (GOOD answers with some thought into them!) will be chosen best answer!:
1. Can anyone give me a real life example of a differential equation that they can workout on as an example for me?
2. How can I solve parametric equations?
3. How do I calculate the instantaneous speed of something without a graph, example of formula and example included?
Thanks!

Answer 1

This will answer questions 1 and 3 in one:

Let a particle be moving through the xy-plane, passing through the point (3,1) with its speed given by

[dy/dx]=3x/y. Find the curve that describes the position of the particle at any x-value.

To solve, first separate the variables, leaving

dy/y = 3x dx and then integrate, giving

ln|y| = (3/2)x^2 + c ------ plug in the initial value statement to solve for c, giving

ln|1| = (3/2)(3^2) + c, 0 = (3/2)(9) + c, c = -(27/2), so

ln|y| = (3/2)x^2 - (27/2), and

y = e^[(3/2)x^2 - (27/2)] is the curve that describes the particle's motion.

For another example, let's say that the rate at which gallons of water, y, flowing into a tank, is modeled by the equation

(dy/dx) + 3y/x= x^2. Find the curve that describes the amount of water in the tank at any given time, given that the initial amount of water at x = 0 is 0 gallons.

You'll notice this equation can't be simply moved around and integrated...rather, we must introduce what is known as an integrating factor - to learn how this is derived, take a differential equations class. At any rate, the integrating factor, I(x) is given by

I(x) = e^(integral[3/x dx]), or more generally, e^(integral[ ]) of the function of x that is adjacent to the y-term on the left side of a differential equation, which in this case, is 3/x. Simplifying gives

I(x) = e^(3ln(x)) = e^(ln(x^3)) = x^3. Now we plug the integrating factor into the equation and solve using a substitution, saying

d/dx[I(x)*y] = I(x)*x^2, so in this case

d/dx[x^3*y] = x^3*x^2 = x^5

Now we integrate with respect to x, giving

x^3*y = (1/6)x^6 + c

at x = 0, y = 0, and we can see that c = 0. so the general equation giving the amount of water in the tank at any x-value is

y = (1/6)x^6 / x^3 = (1/6)x^3



2) As for parametric equations, you simply isolate your common variable, t, in each equation, and then use substitution. For instance,

let x = 2t + 3 and y = t^2 - 3t

Solving for t in the x equation gives t = (x-3)/2. Substituting into the equation for y gives

y = ((x-3)/2)^2 - (3/2)(x-3)

Finally, I'm not sure what exactly you mean here by "without a graph" - if the speed of an object cannot be defined by a function, then you can only make calculate the average speed over a time interval, not the instantaneous speed. For instantaneous speed of an object, take the derivative. For example, let a particle be moving through the xy-plane and its position modeled by the equations

y = 3 - t, x = t^2

to find the instantaneous speed of the particle at a given time, t, find dy/dx:

[dy/dx] = [dy/dt] / [dx/dt] = [-1] / [2t] = -1/(2t)

Answer 2

When a simple electric circuit contains resistance R and inductance L and a voltage E is impressed on it, the current i at any instant in time t is given by the differential equation:
L*di/dt +ri = E.
This can be rewritten as di/dt+(R/L)i = E/L
The integrating factor is e^Rt/L so solution is given by:
i*e^Rt/L = integral E?L e^Rt/L dt =(E/R)*e^Rt/L +C
Thus i = E/R + Ce^-Rt/L If i = 0 when t= 0 and E= Eo, then C= -Eo/R so i = (Eo/R)(1-e^-Rt/L).

If x = sint and y = cos t, then x and y both depend on t so this is a parametric equation.
One way to solve is to eliminate the parameter. Here x^2 = sin^2t and y^2= cos^2t. So x^2+y^2 = sin^2t+cos^t = 1. So x^2+y^2=1 is the equation of the unit circle. Most of the time it is very difficult to eliminate the parameter and so plotting the function is often the way to go just as it is with polar coordinates.

Speed is the magnitude of velocity.
Average speed = change in speed divided by change in time.
The instantaneous speed is the speed at a specific time and is give by ds/dt at t1. Suppose a rectinlinear motion is given by s = 16t^2-48t+80. Then v = ds/dt = 32t-48. If t = t1 = 2, then the instantaneous velocity is 64-48=16.