Curve Fitting Problems -Solutions

   
1.) Given plot of orbital period vs. radius, determine:

A.) Determine the period (in days) and mean orbital radius (in a.u.) of Mercury Mars, and

Jupiter:

Solution: Pick the points off the graph - be careful how you read the logarithmic axes!

Mercury: (88,0.39)

Mars:(690,1.51)

Jupiter:(4200,5.1)

  B.) Eros has an orbital period of 1.8 years. Estimate it's mean distance from the Sun:

Solution: 1.8 years = approx. 657.4 days (365.25 days/yr).

   Draw a line through the datapoints on the graph. Find where that line crosses 657.4 days (just to the left of Mars). Pick off the corresponding orbital radius: about 1.49a.u.

  C.) Determine an appropriate mathematical model for the data represented in this graph:

  SOLUTION: The data is plotted on log-log paper, so we need a power fit, i.e the model will be in the form of:

P = Br^m

It's shown here plotted in linear form on this paper:

y = mx+b where

y = log(P)

x = log(r),  m = m and

b = log(B)

To determine the model, pick 2 points and find the slope using:

m = [(log([y2/y1]))/(log([x2/x1]))]

Using the points for Mercury and Mars:

  m = [(log([690/88]))/(log([1.5/0.39]))] = 1.53

To find B, we can merely pick the y-intercept off the graph. But wait - where is the y-intercept on this graph? On a normal graph, the y-axis is at x=0. But here, we are using a logarithmic axis, where x=log(r), so the y-axis is located at wherever log r=0, or where r =1. On this plot, r=1 a.u. lines up with Earth (which makes sense, since 1 a.u. is defined as the distance from Earth to the Sun). So, our intercept B= about 365

So, our final model is:

P = 365r^1.53

  D.) Neptune is 30.1 a.u. from the sun, what is it's orbital period?

SOLUTION: We'll just plug this value in for the radius in our model above:

P_neptune = 365(30.1)^1.53 = 66,757.4 Days

    2.) Given the following data:

A quick sketch of the data shows that it is clearly not linear:

Sticking the above data in ``Best Fit'' I get:

H = 194.3e^-.157t , which is in fact exponential. I'll replot this on a semi-log scale to make sure:

SOLUTION: We can find this by plugging t=0 into our model, which returns 194 cm.

C.) When will the barrel be empty? An interesting question. Technically never, since

the exponential curve is approaching 0 asymptotically. So, the question is, how empty is empty?

If we assume 99.9% empty, than means (.001)(194cm) =.194 cm,

H = 194.3e^-.157t

LnH = Ln194.3-.157t

t = [(lnH-ln(194.3))/(-.157)] = 44minutes

    3.)Bucket rocks, and spring:

Given the following data:

My plot make a linear fit look reasonable. I ran it through my calculator for confirmation, and

it agreed. The linear model I got was:

H = 1.22-0.0096w

B.) How many pounds of rocks are required to sound the alarm?

SOLUTION: This is the weight when the height is zero,

0.096w = 1.22

w = 1.22/0.0096 = 127.1 lbs

  C.) What is the height when the bucket is empty. Set w=0, you simply get:

H=1.22 meters

    4.) Thermal Runaway:

Given the following data:

Again we see a curve that gets steadily steeper, so I'm guessing a power fit. My calculator agrees,

T = 2.27t^1.49

(where T is temp and t is time)

If 213 degrees is our breakdown point:

log T = log 2.27+1.49log t

log t = [log(213)-log(2.27)]/1.49

log(t) = 1.32

t = 21.07 seconds