Q.NO.4: Find the arc length of the graph of r(t).
r(t)=(t^2)i +(cost + t sint)j +(sint- t cost)k , 0≤t≤π
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s = ∫[0,π] √((2t)^2+(t cost)^2+(t sint)^2) dt
= ∫[0,π] √(4t^2+t^2) dt = √5/2 π^2
To find the arc length of the graph of the vector function r(t) = (t^2)i + (cost + tsint)j + (sint - tcost)k over the interval 0 ≤ t ≤ π, 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 over the interval a ≤ t ≤ b is given by:
L = ∫[a,b] √[f'(t)^2 + g'(t)^2 + h'(t)^2] dt
Let's calculate the derivative of the given vector function, r(t):
r'(t) = (2t)i + (-sint + tsint + cost + tcost)j + (cost - tcost + sint - tsint)k
Now, we can calculate the magnitude of the derivative:
|r'(t)| = √[(2t)^2 + (-sint + tsint + cost + tcost)^2 + (cost - tcost + sint - tsint)^2]
= √[4t^2 + sin^2t + 2tsintcost + 2tsintcost + cos^2t + t^2sin^2t + 2t^2cost + 2t^2sintcost + cos^2t + t^2sin^2t + 2t^2sintcost + 2t^2sintcost + sin^2t + cos^2t + t^2sin^2t]
Simplifying further:
|r'(t)| = √[6t^2 + 4tsintcost + 2sin^2t + 2cos^2t + 2t^2sin^2t + 2t^2cost + 2t^2sintcost]
Now, we can use this magnitude and integrate it over the interval 0 ≤ t ≤ π:
L = ∫[0,π] √[6t^2 + 4tsintcost + 2sin^2t + 2cos^2t + 2t^2sin^2t + 2t^2cost + 2t^2sintcost] dt
Unfortunately, this integral does not have a simple closed-form solution. You would need to use numerical techniques or software to approximate the value of this integral.