AP Calculus AB / Mr. Hansen
2/1/2005 [rev. 2/2/2005]

Name: _________________________

Key to Practice Test on Chapter 6

 

WARNING: This practice test, although it covers a fair number of topics, does not fully prepare you for the real test. For example, there are no problems here on u substitutions, nor are there any examples of solving oddball diffeqs. from scratch (with or without initial conditions). You will want to take a look at the suggested review problems in the 2/2/05 calendar entry if you have not already done so.

 

 

1.

diffeq.
contains a derivative
family of functions or relations
particular
function
relation
differential equation
satisfies
initial conditions

 

 

2.(a)

By inspection, y = 5.28e.074t [no work needed].

(b)

exponential growth

(c)

B
[By the Rule of 72, doubling time is approx. 72/7.4 » 10 time units. You can easily do this without a calculator. On the AP exam, you are not supposed to use the Rule of 72, but you may find it helpful as a sanity check to make sure that you are not selecting a wildly unsuitable answer.]

 

 

3.

ò u dv = uv – ò v du

 

 

4.

Let u = ln 2x, dv = 3x dx. Then du = 1/x dx, and v = 3x2/2 = 1.5x2.
ò 3x ln 2x dx = ò u dv
                      = uv – ò v du
                      = 1.5x2 ln 2x – ò 1.5x dx
                      = 1.5x2 ln 2x – 3x2/4 + C

Before proceeding, verify that d/dx [1.5x2 ln 2x – 3x2/4 + C] equals the integrand.

 

 

5.

Make sure that you cross your z’s for full credit.

d/dz [z ln z – z + C] = [z(1/z) + 1(ln z)] – 1 + 0 = 1 + ln z – 1 + 0 = ln z ü

 

 

6.

General solution is found by inspection, or by recalling what was in your book. The general solution is

T – Tambient = (T0 – Tambient)e–kt
T = Tambient + (T0 – Tambient)e–kt

At t = 30 minutes, we plug in to the solution to get

35 = 30 + 50e–30k
5 = 50e–30k
0.1 = e–30k
ln 0.1 = –30k
–ln 10 = –30k
ln 10 = 30k
k = (ln 10)/30

The units of k are min–1, so as to make the e–kt factor be dimensionless. Don’t worry about that detail for tomorrow’s test. Normally, the only units that I would make a big deal over would be the units of the independent variable (t), the dependent variable (T), or the rate of change (dT/dt), which are minutes, ° C., and ° C./min, respectively.

 

 

7.(a)

If you were unable to sketch the slope field, then use your SLOPES program to display it on your calculator.

(b)

The particular solution is an exponential decay curve (“L-shaped”) passing through (1, 5). If you need to see the curve sketched, please see me before the test and I’ll sketch it for you on paper. Although you were not allowed to use a calculator during the practice test, you could certainly use a calculator now to run the EULER program to see what the curve would look like.

(c)

Dt = –0.1
t0 = 1, M0 = 5, M ′(1) = –M/3 = –5/3
     By Euler’s Method, M1 estimate = M0 + (slope)Dt = 5 + (–5/3)(–1/10) = 5 + 5/30 = 31/6
t1 = 0.9, M1 = 31/6, M ′(0.9) = –M/3 = –(31/6)/3 = –31/18
     By Euler’s Method, M2 estimate = M1 + (slope)Dt = 31/6 + (–31/18)(–1/10) = 31/6 + 31/180

Since you were not allowed to use a calculator, you may leave your answer in either of the two underlined versions. Remember, M2 is an estimate for the value of M(0.8), since we have made two Euler steps of size –0.1. If you wish to check your work with EULER2, the underlined answers above turn out to be 5.339 to the nearest thousandth.

 

 

(d)

Too low, since the Euler track is attempting to use piecewise linear functions in a region where the exact solution has positive concavity. In other words, the true function will always be curving up and away from the straight line segments that Euler’s Method employs.