Calculus II Class 01/13
Calculus I Quiz 0113
Write the equation that says that the rate at which the level y of water in a container changes is proportional to the square root of the depth y of the water.
The rate at which depth changes is y ' , or to more explicitly specify the variables involved, dy / dt.
A quantity is proportional to another if it is equal to a constant multiple of the other. Thus our equation is
y ' = k sqrt(y) or
dy/dt = k sqrt(y)
Write the equation that says that the rate at which velocity v changes is equal to the constant g. Find the general solution to this equation.
The rate at which velocity changes is expressed as dv/dt, so our equation is just
dv/dt = g.
v is the general antiderivative of g with respect to t, so the general solution is
v = g t + c.
Write the equation that says that the rate at which position x changes is equal to the general solution to the preceding equation.
dx/dt = g t + c
x will be the general antiderivative of g t + c, which is 1/2 * g t^2 + c t + d, using d for the new integration constant. (note that we cain't use c because the constants are both arbitrary and are not related). So our solution is
x = 1/2 g t^2 + c t + d.
If we know that when t = 0, v = 0 and x = 0 then what are the velocity and position functions corresponding to constant acceleration g.
Knowing that when t = 0 we have x = 0 tells us that
0 = 1/2 g * 0^2 + c * 0 + d or
d = 0.
So our function is
x = 1/2 g t^2 + c t.
Since v = dx / dt we have v = g t + c, so v = 0 when t = 0 tells us that
0 = g * 0 + c so that
c = 0 .
So our function is just
x = 1/2 g t^2.
If we know that when t = 0, v = 10 and x = 30 then what are the velocity and position functions corresponding to constant acceleration g = - 9.8?
Substituting t = 0, x = 30 into x = 1/2 g t^2 + c t + d we obtain
30 = 1/2 g * 0^2 + c * 0 + d or
d = 30.
So the equation is
x = 1/2 g t^2 + c t + 30.
Now again v = dx / dt = g t + c so substituting t = 0, v = 10 gives us
10 = g * 0 + c or
c = 10.
Thus the equation is
x = 1/2 g t^2 + 10 t + 30.
Now substituting g = -9.8 we have
x = -4.9 t^2 + 10 t + 30.
I just gave a coin an initial velocity of 1.87 m/s from a height of 1.2 meters. Its acceleration is - 9.8 m/s^2. What is its equation of motion, and how can we use this equation to find the time required to reach the floor and its speed at that point? What about its maximum altitude?
We know that dv/dt = accel = -9.8 m/s^2, so the preceding series of problems, using y for position instead of x, gives us the equation
y = 1/2 * g t^2 + c t + d or, using -9.8 for g,
y = -4.9 t^2 + c t + d.
The associated velocity function is
v = dy/dt = -9.8 t + c.
If we start the clock at the instant the coin is flipped, then we know that when t = 0 we have v = 1.87 and y = 1.2.
Substituting t = 0 and y = 1.2 into y = -4.9 t^2 + c t + d gives us d = 1.2
Substituting t = 0 and v = 1.87 into v = -9.8 t + c gives us c = 1.87.
Thus our equation y = -4.9 t^2 + c t + d becomes
y = -4.9 t^2 + 1.87 t + 1.2.
To find when the coin hits the floor we observe that when the coin hits the floor we have y = 0. To find t we therefore solve
0 = -4.9 t^2 + 1.87 t + 1.2.
From the quadratic formula we obtain
t = 0.72 or t = -0.34.
Knowing that the coin hits after we flip it, not before, we use the solution t = 0.72, declaring the solution t = -0.34 invalid since we know that the coin didn't come up through the floor at t = -0.34 s.
The speed of the coin at this instant is found by substituting t = 0.72 into the velocity expression v = -9.8 t + 1.87 to obtain
v = -9.8 * .72 + 1.87 = -5.19,
corresponding to downward velocity 5.19 m/s at the instant the coin hits the floor.
To find the instant when altitude is maximum we can simply maximize the function y = -4.9 t^2 + 1.87 t + 1.2 by finding critical points and using 1st or 2d derivative test.
dy/dt = -9.8 t + 1.87, which gives us a critical point when
-9.8 t + 1.87 = 0 or
t = 1.87 / 9.8 = .19 approx.
Since the second derivative is y '' = -9.8 the second-derivative test tells us that we have a maximum when t = .19.
To get the altitude at that instant we substitute t = .19 into the equation of motion, obtaining
y = -4.9 * .19^2 + 1.87 * .19 + 1.2 = 1.37.
Note that the procedure for maximizing y is equivalent to noting that altitude is maximized when v = 0, since v = 0 is equivalent to -9.8 t + 1.87 = 0.