1. Find, analytically (using derivatives) the points t where the highest and lowest point on the parametric curve
x = x(t) = t + 2sin(2 t),
y = y(t) = t + 2cos(5t), 0 £ t £ p
occur
Solution
: The highest and lowest values of the curve correspond to the maximum and minimum of y(t):
By inspection, the minimum occurs for t around 0.6; the maximum for t approximately 2.6.
Going on to the analysis:
|
y'(t) = 1 -10sin(5*t) = 0 Þ |
|
sin(5t) = 1/10 Þ |
|
5t = arcsin(1/10) + 2nPi, n = 0, 1, 2, .. Þ |
|
t = 0.2 arcsin(0.1) + 0.4*n*Pi, n = 0, 1, 2, .. |
We have to find the values of t that fall in the interval [0, Pi].
There are two angles a and b in [0, Pi] for which sin(a ) = sin(b ) = 1/10. They are a = .1001674212 and b = Pi- a = 3.041425233.
The values of
t = 0.2*.1001674212 + 0.4*n*Pi, n = 0, 1
that fall in the interval [0, Pi] are
.02003348424, 1.276670546, 2.533307607.
The last, t = 2.533307607 corresponds to the maximum of y(t). The maximum value is y(2.533307607) = 4.523282480.
The values of
t = 0.2*3.041425233 + 0.4*n*Pi, n = 0, 1
that fall in the interval [0, Pi] are
.6082850466, 1.864922109, 3.121559170.
The first of these, t = .6082850466 corresponds to the minimum value of y(t). This minimum value is y(6082850466) = -1.381689827.