Exam 1 Corrections Sample

Anonymous Student
Exam 1 Test Corrections
September 25, 2006
Calculus 2
Problem 1: 14/14 Points
Problem 2: 13.5/14 Points
-.5 Points:
In problem 2, section b, I had substituted (1/∞) in for (1/b) and gave the
explanation: “(1/∞) gets infinitely smaller, and thus approaches 0. However,
this is not correct. What I should have said is (1/b) gets infinitely small as b
approaches infinity and thus the quantity (1/b) approaches 0.
Problem 3: 14/15 Points
-0 Points:
The first mistake I made on this problem was a confusion of terms. I said,
“the second derivative has an inflection point” when really the graph of the
function had an inflection point because the second derivative equals 0.
-.5 Points
I simply did not relate my explanation to the function well enough because I
did not show in mathematical terms what I was saying. Thus, concerning the
function P(t), as t gets larger, the quantity 30e-.8t becomes an increasingly
small number because e to a negative exponent is really (1/e). The larger t
becomes, the smaller the quantity 30e-.8t becomes. As t goes to infinity, the
quantity approaches 0. Thus, P(t) taken as a whole becomes 80 divided by
(1+0) which simplifies to 80. Then for larger values of t, P(t) ≈ 80.
-.5 Points
P(t) = .05P is a good approximation of the logistic curve when P is small
because the logistic curve grows similar to an exponential function when P is
small. This is because logistic curves grow similarly to exponential function
but as P gets larger, the logistic curve begins to flatten out (reaching its
carrying capacity) while the exponential function grows without bound.
However, as long as P is small, the quantity (1 – P/1800) in the function P(t)
= .05P(1 – P/1800) will be very close to one making the entire functions
growth similar to P(t) = .05P, an exponential function.
Problem 4: 8.5/12 Points
-3.5 Points
I did not know how to properly set up this problem. I made my mistake in
that I did not know how to express the distance in variables. However, I do
now. I wrote the integral as dW = 0∫6 3x2dx. This should be written dW = 0∫6
3xdx because Work = Force x Distance. The force here is 3x and the distance
is dx. The distance is dx because the spring starts at a length of x and is
stretched to a length of x + dx. So the total displacement is (x + dx) – x which
equals dx.
We are using the integral to sum up infinitely small portions of the work,
d(work). Therefore when computed, this integral would yield all the work
needed to stretch the spring across the entire interval.
Problem 5: 18/18 Points
Problem 6: 12/12 Points
-0 Points
At first I made the careless error as to thinking the entire base of the
triangular cross-section was 3 instead of 6 which lead me to write my
explanation using the area of that cross-section as 3/2y when in truth it is
simply 3y. I realized this error before handing in the test and thought I had
corrected it throughout my entire explanation but I had missed one portion at
the beginning which lead to the grader’s confusion of my work.
Page 1 of 1