Respuesta :
The correct question has r(t) = e^t sint i + e^t cost j + √2 e^t k
Answer:
(a) 2e^t
(b) r(t) = (L/2 + 1)sin(ln|L/2 + 1|) i + (L/2 + 1)cos(ln|L/2 + 1|) j + √2(L/2 + 1) k
Step-by-step explanation:
Given r(t) = e^t sint i + e^t cost j + √2 e^t k
(a) To find the arc length of the function, we first differentiate r(t) to obtain r'(t). Doing this, we have
r'(t) = (cost + sint)e^t i + (cost - sint)e^t j + √2 e^t k
Next, we find the magnitude of r'(t). Doing this, we obtain
|r'(t)| = √{[(cost + sint)e^t]² + [(cost - sint)e^t]²+ (√2 e^t)² k}
= √(4e^(2t))
= 2e^t
(b) Now, because t = 0 corresponds with the (0, 1, √2), we have the arc length function, L to be the integral from 0 to t of |r'(v)|dv
L = integral of 2e^v dv from v = 0 to t.
Evaluation the integral, we have
L = 2e^t - 2e^0
= 2e^t - 2
L = 2(e^t - 1)
Let us make t the subject of the formula.
Divide both sides by 2
L/2 = e^t - 1
Adding 1 to both sides
L/2 + 1 = e^t
e^t = L/2 + 1
Taking natural logarithm of both sides
t = ln|L/2 + 1|
Finally, we put t = ln|L/2 + 1| in the original equation, r(t) we were given to have
r(t) = (L/2 + 1)sin(ln|L/2 + 1|) i + (L/2 + 1)cos(ln|L/2 + 1|) j + √2(L/2 + 1) k
In this exercise we have to use the knowledge of arc length of a function, thus we find that:
(a) [tex]2e^t[/tex]
(b)[tex]r(t) = (L/2 + 1)sin(ln|L/2 + 1|) + (L/2 + 1)cos(ln|L/2 + 1|) + \sqrt 2(L/2 + 1)[/tex]
Given the function as:
[tex]r(t) = e^t sin(t) + e^t cos(t) + \sqrt{2 e^t }[/tex]
(a) To find the arc length of the function, we have:
[tex]r'(t) = (cost + sint)e^t + (cost - sint)e^t+ \sqrt2 e^t\\|r'(t)| = \sqrt{[(cost + sint)e^t]^2 + [(cost - sint)e^t]^2+ (\sqrt2 e^t)^2 }\\= \sqrt(4e^{(2t)})\\= 2e^t[/tex]
(b) We have the arc length function, L to be the integral from 0 to t, so:
[tex]L = \int\limits^t_0 {2e^v} \\L = 2e^t - 2e^0\\= 2e^t - 2\\L = 2(e^t - 1)\\L/2 = e^t - 1\\L/2 + 1 = e^t\\e^t = L/2 + 1\\t = ln|L/2 + 1|\\[/tex]
See more about arc length at brainly.com/question/1577784
