Hello...i had a question. I would really appreciate it if someone could help me...tnx in advance!
K my question is Find the T(t) , N(t), B(t) for r(t)=<2sint,2cost,t>; t=pie/4
...which are normal vectors, binomial vector and i am not sure about T(T)...anyhow...i have the formula for
T(t)
T(t)=r'(t)/abs(r'(t))
so far i got
r'(t)=2cos,-2sint,1
abs(r'(t)=squarroot(2cos(pie/4)-2sin(pie/4)+1)
so plugging it in for
T(t)=r'(t)/abs(r'(t))
it would be
T(t)=2cos(pie/4),-2sin(pie/4),1/squarroot(2cos(pie/4)-2sin(pie/4)+1
is that right...so once i find T(t)...i would fidn N(t) and B(t) ...but so far i just want to check if what i have it right so far...plz plz plz help me...tnz once again
so plugging that for T(t)
To find the tangent vector T(t), you correctly used the formula T(t) = r'(t) / ||r'(t)||, where r'(t) is the derivative of the position vector r(t) and ||r'(t)|| is the magnitude of r'(t).
In your case, r(t) = <2sin(t), 2cos(t), t>.
To find r'(t), differentiate each component of r(t) with respect to t:
r'(t) = <d(2sin(t))/dt, d(2cos(t))/dt, dt/dt> = <2cos(t), -2sin(t), 1>
Next, find the magnitude ||r'(t)||:
||r'(t)|| = sqrt((2cos(t))^2 + (-2sin(t))^2 + 1^2) = sqrt(4cos^2(t) + 4sin^2(t) + 1) = sqrt(4(cos^2(t) + sin^2(t)) + 1)
= sqrt(4 + 1) = sqrt(5)
Now, divide r'(t) by ||r'(t)|| to get T(t):
T(t) = <2cos(t), -2sin(t), 1> / sqrt(5)
So, your expression for T(t) is correct: T(t) = <2cos(t), -2sin(t), 1> / sqrt(5).
To find the normal vector N(t), you can differentiate T(t) with respect to t and normalize the result, similar to how you found T(t). The normal vector N(t) represents the curvature of the curve.
To find the binormal vector B(t), you can take the cross product of T(t) and N(t) at each point. The cross product of two vectors gives a vector that is perpendicular to both of them, which represents the twisting or "turning" of the curve.
Let me know if you need help with finding N(t) and B(t) or if you have any further questions!
To find T(t), N(t), and B(t) for the given vector function r(t) = <2sin(t), 2cos(t), t>, where t = π/4, you're correct in using the formula for T(t):
T(t) = r'(t) / |r'(t)|.
Let's calculate each step:
1. Find r'(t):
r'(t) = <2cos(t), -2sin(t), 1>.
2. Calculate |r'(t)|:
|r'(t)| = sqrt((2cos(t))^2 + (-2sin(t))^2 + 1^2)
= sqrt(4cos^2(t) + 4sin^2(t) + 1)
= sqrt(4(cos^2(t) + sin^2(t)) + 1)
= sqrt(4 + 1)
= sqrt(5).
3. Substitute the values into the formula for T(t):
T(t) = <2cos(t), -2sin(t), 1> / sqrt(5).
Thus, T(t) = <2cos(π/4), -2sin(π/4), 1/sqrt(5)>.
To find N(t) and B(t), you can use the following formulas:
N(t) = (T'(t) / |T'(t)|) × T(t)
B(t) = T(t) × N(t),
where T'(t) is the derivative of T(t) with respect to t.
Find T'(t) using the same process as before (differentiating each component with respect to t) and then substitute the values into the formulas for N(t) and B(t).
I hope this helps!