Answer:
, so find the
tangent
and normal at the point (2,4).
slope of the tangent =
At t = 1, m = 4
Tangent Vector: <1, 4> or
any other
vector in component form that indicates its slope is 4.
Normal Vector: <-4,1> or any
other
vector in component form that indicates its slope is -1/4.
These unit vectors are
and
,
respectively.
The ground speed vector is the resultant of the airspeed vector and the windspeed vector.
airspeed = <325 cos 70o, 325 sin 70o>
since
20o east of north is 70o
in standard position.
windspeed = <40 cos 130o, 40 sin 130o>
since 40o west of north is 130o
in standard position.
ground speed = <325 cos 70o + 40
cos
130o, 325 sin 70o + 40 sin 130o
>
= <85.445, 336.042 >
The magnitude of the ground speed is \/(85.4452 +
336.0422)
= 346.735 mph.
(Note: Use the original values in your
calculator,
not the rounded values)
The direction is obtained by finding