Sample Major Quiz
Problem Number 1
Problem: Derive the expression for the derivative of the function y(t) = a t3 at clock time t.
'Derive' means 'obtain from the definitions'.
The derivative of this function is the instantaneous rate of change of y with respect to t (of the dependent variable with respect to the independent; y is the independent, t is the dependent variable; if you graph y vs. t the derivative is the limiting value of rise / run) .
For all these reasons the derivative is
limit {`dt -> 0} ( (y(t+`dt) - y(t) ) / `dt) =
lim {`dt -> 0}( a (t + `dt)^3 - a t^3) / `dt) =
lim {`dt -> 0) (a t^3 + 3 a t^2 `dt + 3 a t `dt^2 + `dt^3 - a t^3) / `dt =
lim {`dt -> 0}((3 a t^2 `dt + 3 a t `dt^2 + `dt^3) / `dt) =
lim {`dt -> 0} (3 a t^2 + 3 a t `dt + `dt^2) =
3 a t^3.
Problem: If the rate of depth change is rate(t) = .042 t + -1.3, then what is the depth function if the depth at clock time t = 0 is 99? How long does it take for the depth to decrease from 82.87249 to 79.44964? What is the average rate which depth changes over this period?
Interpreting just 'If the rate of depth change is rate(t) = .042 t + -1.3':
Implicitly in this context where the function is rate(t), 'rate of depth change' means 'rate of change of depth with respect to t'. If it's not specified, the rate of change is always with respect to the independent variable.
So now we have
rate(t) = .042 t + -1.3
which is rate of change of depth with respect to clock time.
Depth is a function of t, and this is a statement about the rate of change of that depth function. The depth function isn't even named in the problem, but if we're going to talk about it we need to give it a name. We should give that function a name so we can refer to that function. The name or label we use for this function is up to us. We could use y(t), or f(t), or even depth(t).
Let's choose to call the depth function y(t).
We are given information about the rate of change of y with respect to t (rate(t) = .042 t + -1.3).
The rate of change of y with respect to t is y ' (t). So
rate(t) = y ' (t), or
y ' (t) = .042 t + -1.3.
Now we're asked to find the depth function:
y ' (t) = .042 t + -1.3, so y(t) is the function we would take the derivative of to get .042 t + 1.3.
To get .042 t we would take the derivative of .042 * ( t^2 / 2).
To get -1.3 we would take the derivative of -1.3 t.
We also include a constant c, which has derivative 0.
So y(t) = .042 ( t^2 / 2) + -1.3 t + c or in its most simplified form
y(t) = .021 t^2 - 1.3 t + c.
nutshell summary so far: we are given the rate -of-change-of-depth function and and need the depth function. The derivative of the depth fn is the rate-of-change fn so the depth fn is an antiderivative of the rate fn.
Now we evaluate c. If the depth at t = 0 is 99 then
y(0) = 99 = .021 0^2 - 1.3 0 + c so that
c = 99 and our function is
y(t) = .021 t^2 - 1.3 t + 99.
How long does it take for the depth to decrease from 82.9 to 79.4?
From the y(t) we can find the clock time when the depth reaches each value.
To find the clock time when depth is 82.9 plug 82.9 in for y(t):
82.9 = .021 t^2 - 1.3 t + 99
and solve for t.
Similarly we can find the clock time when depth is 79.4.
What is the average rate which depth changes over this period?
'Average rate of change of depth' implicitly means 'ave rate of change of depth with respect to clock time' (see previous note) so we just calculate `dy / `dt
Additional observation
The depth fn is quadratic so the ave rate should be equal to the midpoint rate; find the midpoint clock time and plug into the rate function and you should get the same thing you got for the average rate.
Problem Number 2
Problem: At what average rate does the exponential population function P(t) = 15 * 1.19 ^ t change between clock times 4.1 and 4.10001?
'ave rate of change' means 'ave rate of change of population with respect to clock time', which is change in population / change in clock time.
Plug in 4.1 to get the first population, 4.10001 to get the second and divide by the change .00001 in clock time. Be sure to use enough significant figures that the change in population has a few significant figures.
In terms of P(t) the calculation would be
(P(4.10001) - P(4.1)) / .00001 =
(P0 * 2^(4.10001 k) - P0 * 2^(4.1 k) ) / .00001.
P0 * 2^(4.10001 k) - P0 * 2^(4.1 k) =
P0 * 2^(4.1 k) (2^.00001 k - 1), etc.
Problem: If the weight of a sandpile whose diameter is 4 meters is 140.8 tons, then what function gives the weight in tons of geometrically similar sandpiles as a function of diameter in meters?
Sand occupies volume so y = k x^3, where y is weight and x is diameter.
Solve for k and substitute to get the function y = 2.2 x^3, approx.
Evaluate y(4.0001) - y(4).
The ave rate is ( y (4.0001) - y(4) ) / (4.0001 - 4), in tons / meters; work out the numbers.
This should be very nearly equal to the derivative of the weight function at x = 4. The derivative of y = 2.2 x^3 is y ' = 6.6 x^2; when x = 4 this is about 106, meaning 106 tons / meter.
Problem Number 3
The depth vs. clock time function y = .027 t2 + -1.1 t + 87 indicates the depth y of water in a certain uniform cylinder at clock time t.
midway the clock time is (6 s + 12 s) / 2 = 9 s.
This is clearly asking for the instantaneous rate of depth change, which is y ' (9). You gotta find the y ' function.
What is the function that represents the rate r of depth change at clock time t?
This is the y'(t) function.
shoulda already done it and should note that you already did
If the rate of depth change is given by dy/dt = .068 t + -2.4 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 6 and t = 12?
depth is antiderivative function y(t) = .068 ( t^2 / 2) - 2.4 t + c = .034 t^2 - 2.4 t + c.
when t = 0 depth is .034 * 0^2 - 2.4 * 0 + c, which is equal to 170 so c = 170.
Problem Number 4
The depth of water in a certain nonuniform container is y = .02 t4 + -2.6 t2 + 89, where depth y is in cm when clock time t is in seconds.
caution: be sure to use enough significant figures to see the pattern of change in the average rates.
The rate at which water flows from a certain nonuniform cylinder is given by rate = .02 t3 + -2.6 t cm3 per minute, where t is in minutes. How much do water do you think will flow between clock times t = 17.3 minutes and t = 34.4 minutes?
depth fn is antiderivative of rate fn; find depth fn then find the change in the depth fn
Problem Number 5
Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
Note: grains are exposed on the surface and surface area proportionality is y = k x^2.