Because of a storm, ground controllers instruct the pilot of a plane flying at an altitude of 1.6 miles to make a $ 90^{\circ} $ turn and climb to an altitude of 1.8 miles. The model for the path of the plane during this maneuver is $\mathbf { r } ( t ) = \langle 4 \cos 4 \pi t , 4 \sin 4 \pi t , 1.6 + 1.6 t \rangle , 0 \leq t \leq \frac { 1 } { 8 } \text { where } t \text { is the }$ time in hours and is the distance in miles. Determine the speed of the plane. Round your answer to The nearest integer.
A)50 mi/h
B)16 mi/h
C)54 mi/h
D)34 mi/h
E)79 mi/h

Question

Quizplus · Accepted Answer

The speed of the plane is the magnitude of the velocity vector $\mathbf{v}(t) = \mathbf{r}'(t)$. Differentiating $\mathbf{r}(t)$, we get $\mathbf{v}(t) = \langle -16\pi \sin 4\pi t, 16\pi \cos 4\pi t, 1.6 \rangle$. The magnitude is $\sqrt{(-16\pi \sin 4\pi t)^2 + (16\pi \cos 4\pi t)^2 + (1.6)^2}$. Simplifying, this becomes $\sqrt{(16\pi)^2 + 1.6^2}$, which is approximately 50 mi/h.

Because of a Storm, Ground Controllers Instruct the Pilot of a Plane