Find T(t), N(t) and B(t) for given vector function, r(t)=<3sint,4t,3cost>
We can start by finding the first derivative:
r'(t) = <3cost, 4, -3sint>
Now we normalize r'(t) to find T(t):
||r'(t)|| = √(3cos²(t) + 16 + 9sin²(t))
T(t) = r'(t) / ||r'(t)|| = <3cos(t)/√(3cos²(t) + 16 + 9sin²(t)), 4 / √(3cos²(t) + 16 + 9sin²(t)), -3sin(t)/√(3cos²(t) + 16 + 9sin²(t))>
Next, we take the derivative of T(t) to find N(t):
T'(t) = <-3sin(t)(9sin²(t)-3cos²(t)-16)/[3cos²(t)+16+9sin²(t)]^(3/2), 0, -3cos(t)(9sin²(t)-3cos²(t)-16)/[3cos²(t)+16+9sin²(t)]^(3/2)>
||T'(t)|| = 3[9sin⁴(t)-18sin²(t)cos²(t)-9cos⁴(t)-32sin²(t)-16cos⁻(t)+256]^(1/2) / [3cos²(t)+16+9sin²(t)]^(3/2)
N(t) = T'(t) / ||T'(t)|| = <-sin(t)(9sin²(t)-3cos²(t)-16)/[9sin⁴(t)-18sin²(t)cos²(t)-9cos⁴(t)-32sin²(t)-16cos⁻(t)+256]^(1/2), 0, cos(t)(9sin²(t)-3cos²(t)-16)/[9sin⁴(t)-18sin²(t)cos²(t)-9cos⁴(t)-32sin²(t)-16cos⁻(t)+256]^(1/2)>
Finally, we obtain B(t) by taking the cross product of T(t) and N(t):
B(t) = T(t) x N(t) = <4cos(t), 3sin(t), 0> / [3cos²(t)+16+9sin²(t)]^(1/2)
isnt |r'(t)|=5?
You are correct, I apologize for the mistake.
||r'(t)|| = √(3²cos²(t) + 4² + 3²sin²(t)) = 5
Therefore, we can simplify the expressions for T(t), N(t), and B(t):
T(t) = r'(t) / ||r'(t)|| = <3cos(t)/5, 4/5, -3sin(t)/5>
N(t) = T'(t) / ||T'(t)|| = <-sin(t), 0, -cos(t)>
B(t) = T(t) x N(t) = <-(3/5)sin(t), 0, -(3/5)cos(t)>
Can you simplify T(t) or is it already simplified?
T(t) is already simplified.
is |T'(t)|=3/5?
No, ||T'(t)|| is not equal to 3/5.
We can calculate ||T'(t)|| as follows:
||T'(t)|| = √((-3sin(t)(9sin²(t)-3cos²(t)-16))^2 + (-3cos(t)(9sin²(t)-3cos²(t)-16))^2) / (3cos²(t) + 16 + 9sin²(t))^(3/2)
Simplifying this expression results in:
||T'(t)|| = 3[9sin^4(t) + 18sin^2(t)cos^2(t) + 9cos^4(t) + 32sin^2(t) + 16cos(t) + 256]^(1/2) / [3cos^2(t) + 16 + 9sin^2(t)]^(3/2)
Therefore, ||T'(t)|| is a function of both sin(t) and cos(t) and it is not a constant value.