Find the length of the curve r(t)=sqrt(2)ti + e^tj + e^-tk, 0<=t<=1
Please explain how it happened. Thank you
s = ∫[0,1]√( e^(t) + e^(-t))^2 dt
s = ∫[0,1]( e^(t) + e^(-t)) dt
s = [0,1]( e^(t) - e^(-t))
s = ( e^(0) - e^(-0)) - ( e^(1) - e^(-1))
s = e - e^(-1)
Just review the derivation of arc length in 3 dimensions, as at
http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx
s = ∫√(f'^2 + g'^2 + h'^2) dt
So, for your problem,
f' = √2
g' = e^t
h' = -e^-t
s = ∫[0,1]√(2 + e^(2t) + e^(-2t)) dt
If that looks hard to do, just note that
2 + e^(2t) + e^(-2t) = (e^2 + e^-t)^2
Well, well, "length of the curve," huh? Are we measuring curves now? Alright, brace yourself for some mathematical fun!
To find the length of a curve, we need to integrate the magnitude of its derivative with respect to t. In simpler terms, we need to find the integral of the speed of the curve over the given interval.
Now let's start by finding the derivative of the curve. Since the curve is given as r(t) = √2ti + e^tj + e^-tk, we can easily differentiate each component:
dr(t)/dt = √2i + e^tj - e^-tj.
The speed of the curve, also known as the magnitude of the derivative, is given by:
|dr(t)/dt| = |√2i + e^tj - e^-tj|.
This takes the square root of the sum of the squares of each component. Let's simplify it further:
|dr(t)/dt| = √(√2)^2 + (e^t)^2 + (e^-t)^2
= √2 + e^(2t) + e^(-2t).
Now it's a matter of finding the integral of this expression over the given interval. We can write it as:
Length = ∫[0,1] √2 + e^(2t) + e^(-2t) dt.
Now, you could plug this expression into a calculator or math software to find the numerical value. But where's the fun in that? Why don't you take this amazing and exciting mathematical integral as an opportunity to practice your integration skills? Trust me, it'll be a hoot!
And voila! You've got the length of the curve. So next time you need to measure the length of a curve, remember to bring your sense of humor and a math expression like this.
To find the length of a curve, we use the formula for arc length.
The arc length of a curve r(t) = x(t)i + y(t)j + z(t)k, where a ≤ t ≤ b, is given by the integral:
L = ∫ [a, b] √( (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 ) dt
In this case, we have the curve r(t) = √(2)ti + e^tj + e^(-t)k, with 0 ≤ t ≤ 1.
Now, we need to find the derivative of r(t) with respect to t:
dr/dt = √(2)i + e^tj - e^(-t)k
Next, we calculate the magnitude of the derivative:
∥dr/dt∥ = √( (√(2))^2 + (e^t)^2 + (e^(-t))^2 )
= √(2 + e^(2t) + e^(-2t))
Finally, we can substitute this derivative into the arc length formula and integrate over the given interval:
L = ∫ [0, 1] √(2 + e^(2t) + e^(-2t)) dt
Evaluating this integral will give us the length of the curve. Note that the exact value of the integral may require numerical methods or approximation techniques.