Find the length L of the curve R(t)=−5cos(4t)i+5sin(4t)j+3tk over the interval [1,5] .

Answer 1

L=∫_a^b√([f′(t)]²+[g′(t)]²+[h′(t)]²)dt

For this case:

a=1

b=5

f(t)=-5cos(4t)

f'(t) =20sin(4t)

g(t)=5sin(4t)

g'(t)=20cos(4t)

h(t) = 3t

h'(t) = 3

(f'(t))²+(g'(t))²+h'(t)² = 400sin²(4t)+400cos²(4t)+9 = 400(1)+9 = 409

Integrating

√409t |⁵₁ = √409(5-1) = 4√409