Question

Define: Derivative of a vector function.​

Answer 1

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The derivative of a vector-valued function can be understood to be an rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.

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Answer 2

$\tt \: \[\begin{align} {r}′(t) &= \lim \limits_{\Delta t \to 0} \dfrac{{r}(t+Δt)− {r} (t)}{Δt} \\ \\ \tt \: &= \lim \limits_{\Delta t \to 0} \dfrac{[(3(t+Δt)+4)\,\hat{\mathbf{i}}+((t+Δt)^2−4(t+Δt)+3)\,\hat{\mathbf{j}}]−[(3t+4) \,\hat{\mathbf{i}}+(t^2−4t+3) \,\hat{\mathbf{j}}]}{Δt} \\ \\ \tt \: &= \lim \limits_{\Delta t \to 0} \dfrac{(3t+3Δt+4)\,\hat{\mathbf{i}}−(3t+4) \,\hat{\mathbf{i}}+(t^2+2tΔt+(Δt)^2−4t−4Δt+3) \,\hat{\mathbf{j}}−(t^2−4t+3)\,\hat{\mathbf{j}}}{Δt} \\ \\ \tt \: &= \lim \limits_{\Delta t \to 0} \dfrac{(3Δt)\,\hat{\mathbf{i}}+(2tΔt+(Δt)^2−4Δt)\,\hat{\mathbf{j}}}{Δt} \\ \\ \tt \: &= \lim \limits_{\Delta t \to 0} (3 \,\hat{\mathbf{i}}+(2t+Δt−4)\,\hat{\mathbf{j}}) \\ \\ \tt \: &=3 \,\hat{\mathbf{i}}+(2t−4) \,\hat{\mathbf{j}} \end{align} \]$

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