3. Find the length of the curve given by r(t)= √(2)ti +e^t j +e^-t k, 0≤t≤1.
ds^2 = t^2 + e^2t + e^-2t = (e^t + e^-t)^2
ds = e^t + e^-t = 2cosh(t)
so the arc length is
∫[0,1] 2cosh(t) dt = 2sinh(t) [0,1] = 2sinh(1) = e - 1/e
ds = e^t + e^-t = 2cosh(t) what this mean? does it mean e^t +e^-t is the same ascosh(t)?
To find the length of the curve given by the vector function r(t), we can use the arc length formula. The arc length formula for a vector function r(t) = f(t)i + g(t)j + h(t)k, where a ≤ t ≤ b is given by:
S = ∫[a, b] √[ (df/dt)^2 + (dg/dt)^2 + (dh/dt)^2 ] dt
In our case, r(t) = √(2)ti + e^t j + e^(-t) k. So, we need to find the derivatives of f(t), g(t), and h(t) first.
df/dt = (√2)t' = √2
dg/dt = e^t
dh/dt = -e^(-t)
Next, we plug these derivatives into the arc length formula and integrate with respect to t over the given interval:
S = ∫[0, 1] √[ (√2)^2 + (e^t)^2 + (-e^(-t))^2 ] dt
= ∫[0, 1] √[2 + e^(2t) + e^(-2t)] dt
To solve this integral, we need to use techniques such as substitution or integration by parts. These techniques might require further simplification and manipulation.
Unfortunately, the solution to this integral involves special functions that cannot be expressed with elementary functions. Thus, we won't be able to find an exact solution to the length of the curve.
However, if numerical approximation is acceptable, you can approximate the length of the curve using numerical integration techniques such as Simpson's rule or the trapezoidal rule. These methods can provide a close approximation of the length of the curve within a desired level of accuracy.