Question

1) If vec{u}(t) = \sin(2t), \cos(6t), t and vec{v}(t) = t, \cos(6t), \sin(2t) , use the formula below to find the given derivative. frac{d}{dt} [\vec{u} (t) \cdot \vec{v}(t)] = \vec{u}'

Answer 1

Answer and Explanation:

To do the computations, we need the derivatives of u and v. We calculate them as:

u′(t)=⟨2cos2t,−6sin6t,1⟩

v′(t)=⟨1,−6sin6t,2cos2t

(A) We compute:

ddt(u(t)⋅v(t))=u′(t)⋅v(t)+u(t)⋅v′(t)

ddt(u(t)⋅v(t))=⟨2cos2t,−6sin6t,1⟩⋅⟨t,cos6t,sin2t⟩+⟨sin2t,cos6t,t⟩⋅⟨1,−6sin6t,2cos2t⟩

ddt(u(t)⋅v(t))=2tcos2t−6sin6tcos6t+sin2t+sin2t−6sin6tcos6t+2tcos2t

ddt(u(t)⋅v(t))=4tcos2t−12sin6tcos6t+2sin2t

(B) We compute:

ddt(u(t)×v(t))=u′(t)×v(t)+u(t)×v′(t)

ddt(u(t)⋅v(t))=⟨2cos2t,−6sin6t,1⟩×⟨t,cos6t,sin2t⟩+⟨sin2t,cos6t,t⟩×⟨1,−6sin6t,2cos2t⟩

ddt(u(t)⋅v(t))=⟨−cos(6t)−6sin(2t)sin(6t),t−2cos(2t)sin(2t),2cos(2t)cos(6t)+6tsin(6t)⟩+⟨−cos(6t)−6sin(2t)sin(6t),t−2cos(2t)sin(2t),2cos(2t)cos(6t)+6tsin(6t)⟩

ddt(u(t)⋅v(t))=⟨−2cos(6t)−12sin(2t)sin(6t),2t−4cos(2t)sin(2t),4cos(2t)cos(6t)+12tsin(6t)⟩