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{\bf EML 5060\hfill Analysis in Mechanical Engineering \hfill Fall 2002 }\\
{\it Test 1 \hfill Van Dommelen (dommelen@eng.fsu.edu) \hfill Due 8/30/02 }
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Hand in the solution to this test on 8/30/02 (5\% of your final
grade).  If your performance is insufficient, you will need to hand in
a corrected version; however, only your initial grade counts.  Please
note: This test must have been {\it accepted} before Exam 1 Calculus,
or you also receive a 0 grade for exam 1. {\em Read carefully. Look it
up.  Answer questions in order from left to right, top to bottom.}

Neatly draw the graph of the following functions, showing the locations
of 0 and $\pm 1$ on each axis. Give the derivative.  Indicate non-principal
values as a broken line.  Make sure that you give enough of the curves
to {\it clearly} demonstrate {\it all} features.
{\it Make sure that you have answered all parts, including derivatives}.
$$
2x-2 \qquad \qquad \qquad \qquad
x^2 + 1 \qquad \qquad \qquad \qquad
x^4 - x^2
$$
\vfill
$$
\sin(x)\qquad \qquad \qquad \qquad
\arcsin(x)\qquad \qquad \qquad \qquad
\sinh(x)
$$
\vfill
$$
\cos(x)\qquad \qquad \qquad \qquad
\arccos(x)\qquad \qquad \qquad \qquad
\cosh(x)
$$
\vfill
$$
\tan(x)\qquad \qquad \qquad \qquad
\arctan(x)\qquad \qquad \qquad \qquad
\tanh(x)
$$
\vfill
$$
\ln(x)\qquad \qquad \qquad \qquad
e^x\qquad \qquad \qquad \qquad
\tan(x^2)
$$
\vfill

Find (include any integration constants and absolute signs):
$$
\int x^{-2} {\rm d} x= \qquad \qquad \qquad
\int_1^2 x^{-2} {\rm d} x = \qquad \qquad \qquad
\int_1^x \xi^{-2} {\rm d} \xi =
$$
$$
\int {{\rm d} x \over x} = \qquad \qquad \qquad
\int {1\over 1 - x^2} {\rm d} x = \qquad \qquad \qquad
\int {1\over 1 + x^2} {\rm d} x =
$$
$$
\int \ln(x) {\rm d} x = \qquad \qquad \qquad
\int x e^x {\rm d} x = \qquad \qquad \qquad
\int x e^{x^2} {\rm d} x =
$$

$$\left| \matrix{
1 & 2 & 3\cr
2 & 3 & 4\cr
3 & 4 & 5}
\right| = \qquad \qquad \qquad
\lim_{x\to 0} {\tan(x) \over x} = \qquad \qquad \qquad
{{\rm d} \over {\rm d} x} \int_x^2x f(\xi) {\rm d}\xi =
$$

$$ 2 + 1 + 0 - 1 -2 -3 -4 \ldots -99 -100  = \qquad \qquad\qquad\qquad\qquad
e^{2} + e^{1} + e^{0} + e^{-1} + e^{-2} + e^{-3} + e^{-4} + \ldots = $$

$$ {\rm Solve: }\quad{{\rm d} y\over {\rm d} x} = y \qquad y(1)=1$$

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