Mathematics

Mathematics

Problem

Find f′(2)f'(2), where f(t)=u(t)⋅v(t)f(t) = \mathbf{u}(t) \cdot \mathbf{v}(t), u(2)=⟨1,2,−1⟩\mathbf{u}(2) = \langle 1, 2, -1 \rangle, u′(2)=⟨3,0,4⟩\mathbf{u}'(2) = \langle 3, 0, 4 \rangle, and v(t)=⟨t,t2,t3⟩\mathbf{v}(t) = \langle t, t^2, t^3 \rangle.
Intuitive explanation

You're working with the product rule for dot products: when f(t)=u(t)⋅v(t)f(t) = \mathbf{u}(t) \cdot \mathbf{v}(t), the derivative is f′(t)=u′(t)⋅v(t)+u(t)⋅v′(t)f'(t) = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t). The key insight is that you can differentiate v(t)\mathbf{v}(t) directly since it's given explicitly, then evaluate both dot products at t=2t = 2 using the values you're provided.

Intuitive explanation

You're working with the product rule for dot products: when f(t)=u(t)⋅v(t)f(t) = \mathbf{u}(t) \cdot \mathbf{v}(t), the derivative is f′(t)=u′(t)⋅v(t)+u(t)⋅v′(t)f'(t) = \mathbf{u}'(t) \cdot \mathbf{v}(t) + \mathbf{u}(t) \cdot \mathbf{v}'(t). The key insight is that you can differentiate v(t)\mathbf{v}(t) directly since it's given explicitly, then evaluate both dot products at t=2t = 2 using the values you're provided.

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