Calculus: The Language of Change

Exercise Set 4.1
The Cool Canary

by Wyatt Hadley

Exercise   1

A Practical Two-Point Formula

What  is the equation of the line that passes through the points (0,75) and (10,60)?

Use your equation to predict the temperature of the canary at 6:00 am, 10 hours after Jonnie does his vicious deed.

Solution  to  the  first  part

The slope of the line is -3/2.  This is found by subtracting the coordinates of the two points, and dividing the difference in 'C' by the difference in 't'.
The 'y' intercept is given in the first point, and is 75.
We can then easily epress an equation for the line in slope-intercept form:
C=-(3/2)t +75

Solution  to  the  second  part

The value of 't' at 6:00 am is 600 min (10hours*60min/hour).
and, C[600]=-(3/2)(600)+75= -825 degrees F!
Better throw this bird back in the oven!

Exercise   2

Tapering  Off is a ?? in The Rate of Cooling

What happens to the rate of cooling on the nonlinear graphs above as the temperature  decreases toward zero?  Does the rate of cooling decrease or increase as the temperature decreases toward zero?
We expect that when C is big, the rate of cooling is big, and that when C is near zero, the rate of cooling is small.  Why do we expect this?  Which graphs in the Figures above have this feature?

Answer

The rate of cooling decreases as the temperature decreases towards zero by observing that the slope of the graph becomes flatter as the temperature decreases.
Most of us would expect that the greater the temperature difference, the faster the rate of temperature change.  For example, if we want to cool off a warm can of soda pop quickly, we might put it in the freezer rather than refrigerator, though the refrigerator would be faster than putting it in our cool basement.  Figures 4.2 and 4.3 graphically illustrate this feature.

Exercise   3

Mindless Fits

Show that all of the functions below satisfy C[0] = 75 and C[10] = 60.

a)  C = (a/(t + k)),  a = 3000,  k = 40

b)  C = (a/(Sqrt[t + k])),  a = 100 (Sqrt[ 10]),  
    k = 160/9

c)  C = (a/(((t + k))^2)),   
    a = 10000 (27 +  12 (Sqrt[ 5])) ≈ 538328,  
    k = 40 + 20 (Sqrt[ 5]) ≈ 84.7214

d)  C = 75  (e^(-kt)),  
    k = Log[5/4]/10 ≈ 0.0223144

Solution   to A

\(Clear[t, c]; 
c[t_]\  := \ 3000\/\(t + 40\); 

    Print["c[0] = ", c[0]]; 
Print["c[10] = ", c[10]]; 

    Plot[c[t]\ , \ {t, 0, 600}]; 
Print["c[600] = ", N[c[600]]]; \)

Solution   to B

\(\(Clear[t, c]; 
c[t_]\  := \(100 \@ 10\)\/\@\(t + 160/9\); 

    Print["c[0] = ", c[0]]; 
Print["c[10] = ", c[10]]; 

    Plot[c[t]\ , \ {t, 0, 600}]; 
Print["c[600] = ", N[c[600]]]; 

    \)\)

Solution   to C

\(\(Clear[t, c]; 

    c[t_]\  := \(10000 \((27 + \ 12 \@ 5)\)\)\/\((t + 40 + 20 \@ 5)\)\^2\ \ ; 
    
Print["c[0] = ", N[c[0]]]; 
Print["c[10] = ", N[c[10]]]; 

    Plot[c[t]\ , \ {t, 0, 600}]; 
Print["c[600] = ", N[c[600]]]; 

    \)\)

Solution   to D

\(\(Clear[t, c]; 
c[t_]\  := 75  Exp[\(-0.1\) Log[1.25] t]; 

    Print["c[0] = ", N[c[0]]]; 
Print["c[10] = ", N[c[10]]]; 

    Plot[c[t]\ , \ {t, 0, 600}]; 
Print["c[600] = ", N[c[600]]]; 

    \)\)