Question

X= t^2, y= sin п t, z= cosп t t= 1 c

Answer 1

Проекции скорости на оси:

Vx(t) = x'(t) = 3·(-sin(pi·t^2/3)·(2pi·t/3) = -2pi·t·sin(pi·t^2/3).

Vy(t) = y'(t)=2pi·t·cos(pi·t^2/3).

V(t)=корень(Vx^2+Vy^2)= 2pi·t.

V(1)=2 pi = 6,28 м/с.

Проекции ускорения на оси:

ах(t)=Vx'(t)= -2pi·(sin(pi·t^2/3)+2pi·t/3 · t· cos(pi·t^2/3)).

ay(t)=Vy'(t)= 2pi·(cos(pi·t^2/3)-2pi·t^2/3 · sin(pi·t^2/3).

ax(1)=-2pi(sin(pi/3)+2pi/3·cos(pi/3))= -2pi( ¥3/2 + pi/3)= -pi/3 ·(3¥3 + 2pi).

ay(1)=2pi(cos(pi/3)-2pi/3·sin(pi/3)=pi/3·(3-2pi¥3).

a= корень(ах^2+ау^2)= pi/3·корень(27+4pi^2 + 12pi¥3 + 9 + 12pi^2-12pi¥3)=pi/3 корень(36+16pi^2).

Касательное ускорение:

at=V'(t)=(2pi·t)'=2pi.

Нормальное ускорение:

an= корень(а^2 - ат^2)= корень(-4pi^2+36pi^2/9 +16pi^4/9)= 4pi^2/3=4·3,14·3,14/3= 13,16 м/с^2.

Answer 2

To evaluate the values of x, y, and z at t = 1, we can substitute t = 1 into the given expressions:

x = t^2 x = 1^2 x = 1

y = sin(пt) y = sin(п * 1) y = sin(п) y = 0

z = cos(пt) z = cos(п * 1) z = cos(п) z = -1

Therefore, when t = 1, x = 1, y = 0, and z = -1.