Answer to Question #99028 in Calculus for pei

Question #99028
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
1
Expert's answer
2019-11-20T12:58:06-0500

Speed of the Plane


Given,

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


r(t)=10cos10π i+10sin10πt j+(4+4t)k, 0t120r(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)||r' (t)||


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



v(t)=r(t)=ddt(10cos10π i+10sin10πt j+(4+4t)k)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 )

=100πsin10π t i + 100πcos10πt j+4k= - 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)=100πsin10π t i + 100πcos10πt j+4k ||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 ||


=(100πsin10πt)2+(100πcos10πt)2+42= \sqrt {(- 100 \pi sin10\pi t)^2 + (100\pi cos 10\pi t)^2 + 4^2 }=(10000π2sin210πt)+(10000π2cos210πt)+16= \sqrt {( 10000 \pi^2 sin^2 10\pi t) + (10000\pi^2 cos^2 10\pi t) + 16 }


=10000π2 [(sin2(10πt)+cos2(10πt)]+16= \sqrt {10000 \pi^2 \space [( sin^2 (10\pi t ) + cos^2 (10\pi t)] + 16}



v(t)=10000π2(1)+16=10000π2+16||v(t) || = \sqrt {10000\pi^2 (1) + 16} = \sqrt {10000\pi^2 + 16}

Now the Speed of the plane==4×625π2+1=314.025 miles/hrNow \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 αT and αNα_{T} \space and \space α_{N}


Here, formula for Tangential acceleration is given by


αT=ddt[v]=v.αvα_{T} = \frac {d}{dt} [||v||] = \frac {v.α}{||v||}

We know,

α(t)=v(t)=ddt(100πsin10π t i + 100πcos10πt j+4k)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 )

=1000π2 cos10πti1000π2sin10πt j= -1000 \pi^2 \space cos 10\pi t \vec i - 1000 \pi^2 sin 10\pi t \space \vec j



v.α=(100πsin10π t i + 100πcos10πt j+4k).(1000π2 cos10πti1000π2sin10πt 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)


=100000π3sin(10πt) cos(10πt)100000π3sin(10πt)cos(10πt)+0= 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=0v.α = 0


So, Tangential acceleration is

αT=v.αv=0314.025=0α_{T} = \frac {v.α}{||v||} = \frac {0} { 314.025} = 0

Here,  formula for Normal components of acceleration is given by


αN=v×αvα_{N} = \frac {||v \times α||}{||v||}


v×α=ijk100πsin(10πt)100πcos(10πt)41000π2cos(10πt)1000π2sin(10πt)0v \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+4000π2sin(10πt))j(0+4000π2cos(10πt)+k(100000π3sin2(10πt+100000π3cos2(10πt))= 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×α==i(4000π2sin(10πt))j(4000π2cos(10πt)+k×100000π3(sin2(10πt+cos2(10πt)))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×α=i(4000π2sin(10πt))j(4000π2cos(10πt)+k×100000π3v \times α = i ( 4000 \pi^2 sin (10\pi t)) - j ( 4000\pi^2 cos (10\pi t) \\+ k \times 100000 \pi^3


v×α=(4000π2)2sin2(10πt)+(4000π2)2cos2(10πt)+100002π6|| 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×α=(4000π2)2 [sin2(10πt)+cos2(10πt)]+100002π6|| v \times α ||= \sqrt {(4000 \pi^2)^2 \space [sin ^2(10\pi t) + cos^2 (10\pi t)] + 10000^2 \pi^6}

v×α=(4000π2)2 +100002π6|| v \times α ||= \sqrt {(4000 \pi^2)^2 \space + 10000^2 \pi^6}



v×α=4000π21+625π2|| v \times α ||= 4000 \pi^2 \sqrt {1 +625 \pi^2}

αN=v×αv=4000π21+625π24×625π2+1α_{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

αN=1000π2α_{N} = 1000 \pi^2



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