For the given position vectors r ( t )r(t) compute the unit tangent vector T ( t

Question

For the given position vectors r(t)r(t) compute the unit tangent vector T(t)T(t) for the given value of tt

A) Let r(t)=(cos3t,sin3t)Let r(t)=(cos3t,sin3t).

Then T(/4)=(_,_,_)

B) Let r(t)=(t^2,t^3)
Then T(3)= (_,_,_)

C) Let r(t)=e^3ti+e^3tj+tk

Then T(-1) = (__,__,__)

ALSO

Find a vector parametric equation r (t)r(t) for the line through the points P=(4,1,1)P=(4,1,1) and Q=(3,2,4)Q=(3,2,4) for each of the given conditions on the parameter tt.

f r (0)=4,1,1r(0)=4,1,1 and r (8)=3,2,4r(8)=3,2,4, then
r (t)=

(b) If r (7)=Pr(7)=P and r (10)=Qr(10)=Q, then
r (t)=

(c) If the points PP and QQ correspond to the parameter values t=0t=0 and t=3t=3, respectively, then
r (t)=r(t)=

Explanation / Answer

T(t) = v(t)/||v(t)|| = r'(t)/||r'(t)||.

1) Let r(t) = <cos3t, sin3t>

r'(t) = <-3sin3t, 3cos3t> = v(t)

||v(t)|| = sqrt(9(sin3t)^2 + 9(cos3t)^2) = sqrt(9) = 3

T(t) = <-sin(3t), cos(3t)>

T(pi/4) = <-sin(3pi/4), cos(3pi/4)>

2) Let r(t) = <t^2, t^3>

r'(t) = <2t, 3t^2> = v(t)

||v(t)|| = sqrt(4t^2 + 9t^4) = tsqrt(4 + 9t^2)

T(t) = <2/sqrt(4 + 9t^2), 3t/sqrt(4 + 9t^2)>

T(3) = <2/sqrt(85), 9/sqrt(85)>

3) Let r(t) = <e^3t, e^-3t, t>

r'(t) = <3e^3t, -3e^-3t, 1> = v(t)

||v(t)|| = sqrt(9e^(6t) + 9e^(-6t) + 1)

T(-1) = <((3e^-3)/sqrt(1+9(e^6t+e^-6t))), (-3e^-3t/(sqrt(1+9(e^6t+e^-6t))), (1/sqrt(1+9(e^6t+e^-6t)))>

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects