Answer to Question #99028 in Calculus for pei

Question

Because of a storm, ground controllers instruct the pilot of a plane flying at an altitude of 4 miles to make a turn and climb to an altitude of 4.2 miles. The model for the path of the plane during this maneuver is
r(t) = 10 cos 10π t i + 10 sin 10 π t j + ( 4+ 4t )k , 0≤ t≤ 1/20
where t is the time in hours and r is the position vector.
i. Determine the speed of the plane
ii. Calculate αT and αN

Answer 1

Speed of the Plane

Given,

The model for the path of the plane during this maneuver is

$$r(t) = 10 cos 10\pi \space \vec i + 10 sin 10\pi t \space \vec j + (4 + 4t ) \vec k , \space 0\le t\le\frac {1}{20}$$

Here t is the time in hours and

r is the position vector

(a)

We can calculate the speed of the plane by the formula

$$||r' (t)||$$

So, Differentiate the position vector r(t) with respect to t

$$v(t) = r' (t) = \frac {d}{dt} (10 cos 10\pi \space \vec i + 10 sin 10\pi t \space \vec j + (4 + 4t ) \vec k )$$

$= - 100\pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k$

Now the Speed of the plane =

$$||v(t) || = ||r' (t) || = || - 100 \pi sin 10\pi \space t \space \vec i \space + \space 100\pi cos 10\pi t \space \vec j + 4 \vec k \space ||$$

$$= \sqrt {(- 100 \pi sin10\pi t)^2 + (100\pi cos 10\pi t)^2 + 4^2 }$$ $$= \sqrt {( 10000 \pi^2 sin^2 10\pi t) + (10000\pi^2 cos^2 10\pi t) + 16 }$$

$$= \sqrt {10000 \pi^2 \space [( sin^2 (10\pi t ) + cos^2 (10\pi t)] + 16}$$

$$||v(t) || = \sqrt {10000\pi^2 (1) + 16} = \sqrt {10000\pi^2 + 16}$$

$$Now \space the \space Speed \space of \space the \space plane = = 4 \times \sqrt {625\pi^2 + 1} = 314.025 \space miles/hr$$

(b).

Now, we need to calculate the $α_{T} \space and \space α_{N}$

Here, formula for Tangential acceleration is given by

$$α_{T} = \frac {d}{dt} [||v||] = \frac {v.α}{||v||}$$

We know,

$$α(t) = v'(t) = \frac {d}{dt} (- 100 \pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k )$$

$= -1000 \pi^2 \space cos 10\pi t \vec i - 1000 \pi^2 sin 10\pi t \space \vec j$

$$v.α = (- 100\pi sin 10\pi \space t \space \vec i \space + \space 100 \pi cos 10\pi t \space \vec j + 4 \vec k ). (-1000 \pi^2 \space cos 10\pi t \vec i - 1000 \pi^2 sin 10\pi t \space \vec j)$$

$= 100000\pi^3 sin (10\pi t) \space cos (10\pi t) - 100000 \pi^3 sin (10\pi t) cos (10\pi t) + 0$

$v.α = 0$

So, Tangential acceleration is

$$α_{T} = \frac {v.α}{||v||} = \frac {0} { 314.025} = 0$$

Here,  formula for Normal components of acceleration is given by

$$α_{N} = \frac {||v \times α||}{||v||}$$

$$v \times α = \begin{vmatrix} i & j & k \\ -100 \pi sin (10\pi t) & 100\pi cos (10\pi t) & 4 \\ -1000\pi^2 cos (10\pi t) & -1000 \pi^2 sin (10\pi t) & 0 \end{vmatrix}$$

$$= i (0 + 4000 \pi^2 sin (10\pi t)) - j (0 + 4000\pi^2 cos (10\pi t) \\+ k ( 100000 \pi^3 sin^2 (10\pi t + 100000 \pi^3 cos^2 (10\pi t))$$

.

$$v \times α = = i ( 4000 \pi^2 sin (10\pi t)) - j ( 4000\pi^2 cos (10\pi t) \\+ k \times 100000 \pi^3 (sin^2 (10\pi t + cos^2 (10\pi t)))$$

$$v \times α = i ( 4000 \pi^2 sin (10\pi t)) - j ( 4000\pi^2 cos (10\pi t) \\+ k \times 100000 \pi^3$$

$$|| v \times α ||= \sqrt {(4000 \pi^2)^2 sin ^2(10\pi t) + (4000 \pi^2)^2 cos^2 (10\pi t) + 10000^2 \pi^6}$$

$$|| v \times α ||= \sqrt {(4000 \pi^2)^2 \space [sin ^2(10\pi t) + cos^2 (10\pi t)] + 10000^2 \pi^6}$$

$$|| v \times α ||= \sqrt {(4000 \pi^2)^2 \space + 10000^2 \pi^6}$$

$$|| v \times α ||= 4000 \pi^2 \sqrt {1 +625 \pi^2}$$

$$α_{N} = \frac {||v \times α||}{||v||} = \frac {4000\pi^2 \sqrt {1+ 625\pi^2}}{4 \times \sqrt {625\pi^2 + 1}}$$

Normal components of acceleration is

$$α_{N} = 1000 \pi^2$$