The motion of a particle is described by the set of parametric equations $x=\frac{t}{5}+1$, $y=2t-e^{\frac{t}{2}}+1$, for $0\le t\le 5$
a) Find the coordinates of the absolute maximum and minimum. Justify your answer $\frac{dy}{dt}=2-\frac{1}{2}e^{\frac{t}{2}}=0$
$t=2.77259$
$\begin{array}{|c|c|c|} \hline t & x(t) & y(t)\\ \hline 0 & 1 & 0\\ \hline 2.77259 & 1.55452 & 2.54518\\ \hline 5 & 2 & -1.18249\\ \hline \end{array}$
Answer: The absolute max is 2.545 and occurs at t=2.773, and the absolute min is -1.182 and occurs at t=5 by EVT.
b) Find the x-intercept of the line tangent to the graph at t=2.
$\frac{dx}{dt}=\frac{1}{5}$
$\frac{dy}{dx}$ when t=2 is 3.204245
x(2)=1.4; y(2)=2.282
y-2.282 = 3.204(x-1.4)
Answer: The x-intercept of the line tangent to the graph at t=2 is 1.4.
c) Find the length of the graph from t=0 to t=5
$\int _0^5\sqrt{\left(\frac{1}{5}\right)^2+\left(2-\frac{1}{2}e^{\frac{t}{2}}\right)^2dt}=6.426\:$