How do you work T = .2 sqrt(L) backwards?

We can solve the equation for L if we first multiply both sides by 1/.2 then square both sides:

T = .2 sqrt(L) so

T / .2 = sqrt(L); squaring both sides we have

T^2 / .04 = L.  Switching both sides and simplifying:

L = 25 T^2.

To get to frequency remember that frequency in cycles/minute it 60 / T, where T is period in seconds.  Thus f = 60 / (.2 sqrt(L) ) = 300 / sqrt(L).

Quote:  'I understand that you take the dependent variable and divide it by the independent variable to get average rate of change'.

This is an almost correct statement.  But it should be

'You take the change in the dependent variable and divide it by the change in the independent variable to get average rate of change'.  The key is 'change in'.

For example the average rate of change of frequency with respect to pendulum length between lengths L = 36 cm and L = 81 cm.  Frequency is f = 300 / sqrt(L) so

• if L = 36 cm then f = 300 / sqrt(36) = 50 cycles/min
• if l = 81 cm then f = 300 / sqrt(81) = 33.333... cycles/min

We need to find change in frequency divided by change in length:

• change in frequency is 33.33..  - 50 = -16.66.. cycles/min
• change in length is 81 cm - 36 cm = 45 cm.
• ave rate of change is therefore

ave rate = (-16.66... cycles/min ) / (45 cm) = -.38 = -.38 (cycles/min)/cm.

Based on this result approximately how much change would you expect in a pendulum of length 50 cm if you changed the length by 5 cm?  Why is this only an approximation and not an exact result?

How do you make a trapezoidal approximation graph?

Suppose you have positions y = 4 cm, 10 cm, 14 cm and 16 cm at clock times t = 10 s, 20 s, 30 s, 40 s.

Construct the trapezoidal graph by constructing vertical line segments starting from the t-axis points and rising to altitudes representing the respective positions y.

The changes in y over the three intervals are 6 cm, 4 cm and 2 cm.  The changes in clock time are 10 sec on each interval.  Average rates of change on the three intervals are therefore

• first interval:  ave rate = 6 cm / 10 sec = .6 cm/sec.
• second interval:  ave rate = 4 cm / 10 sec = .4 cm/sec.
• third interval:  ave rate = 2 cm / 10 sec = .2 cm/sec.

Rises on the graph represent changes in y, runs represent changes in t.

These average rates therefore represent rises / runs, or slopes.

We label the slopes using rectangular boxes.

How do we divide 1 by 43 to 42 significant figures?

Use long division.  It only takes a few minutes.  Or look for a way to get a calculator with only 9 or 12 or whatever number of significant figures to still help with the process.

To find the volume of the cylindrical portion of the can using paper rulers what do we do?

Suggestions:

• measure the height
• measure the radius
• wait, we don't know the exact center point, so measure the diameter.
• we'd have to cut the thing to measure diameter so measure circumference
• pi is the ratio of circumference to diameter.  That is, you have pi diameters in a circumference.  So we can find the diameter from the circumference.
• in order to find the volume we need the area of the cross-section to go with the height of the can; the product of cross-sectional area and altitude is volume
• the cross-section is a circle, whose area is pi r^2.  We can find diameter from circumference so we can find the radius.

Suppose the cylindrical portion of the can has radius 3 cm.  If the water level in the can goes down 2 cm in one gulp, then how many cubic cm of badly flavored sugarwater did you just consume?

• Cross-sectional area is pi r^2 = pi ( 3 cm)^2 = 9 pi cm^2.
• height of what was consumed is 2 cm
• so volume is c.s. area * height = 9 pi cm^2 * 2 cm = 18 pi cm^3, or about 56.6 cm^3(?).

If the 18 pi cm^3 flows out into a tube whose cross-sectional area is .3 pi cm^2 then how much of the tube is filled?

• The tube will form a column with cs area .3 pi cm^2.  Volume is 18 pi cm^3.
• We can write an equation in terms of height h in the tube.  Volume in tube = c.s. area of tube * h so

18 pi cm^3 = .3 pi cm^2 * h.

Solve (divide both sides by .3 pi cm^2):

h = 18 pi cm^3 / (.3 pi cm^2) = 60 cm.

If the depth of sugarwater in the can decreases from 6 cm to 5.4 cm in 10 seconds then at what average rate is sugarwater being sucked from and/or leaking out of the can?

If sugarwater depths at t = 0, 10, 20 and 30 seconds are respectively 6, 5.4, 4.9 and 4.5 cm then at what average rate is it being removed from the can during each interval?

After taping the paper ruler to the bottle, fill it and then time depth vs. clock time at regular intervals.  use a pendulum to do the timing.  Try to get five intervals between start and the time the depth above the outflow is about 2 cm.

Find the volume of the cylindrical section of the bottle.

Measuring only while water is in the cylindrical section, calculate average rate of depth change over each interval and construct a labeled trapezoidal approximation graph for depth vs. clock time.

If possible construct a graph of rate of flow in cm^3 / sec for these intervals.

Finish anything you can and pose any new questions about previous assignments.