1. The problem is to define the derivative of vector-valued functions and understand how to compute it with examples. 2. A vector-valued function is a function that outputs a vector for each input, typically written as $$\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle$$ where $x(t)$, $y(t)$, and $z(t)$ are scalar functions of $t$. 3. The derivative of a vector-valued function with respect to $t$ is defined as the vector of derivatives of its components: $$\mathbf{r}'(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right\rangle$$ This represents the rate of change of the vector function at each point $t$. 4. Important rule: The derivative is taken component-wise, meaning you differentiate each component function separately. 5. Example 1: Let $$\mathbf{r}(t) = \langle t^2, \sin(t), e^t \rangle$$ Compute $$\mathbf{r}'(t)$$. 6. Differentiating each component: - $$\frac{d}{dt} t^2 = 2t$$ - $$\frac{d}{dt} \sin(t) = \cos(t)$$ - $$\frac{d}{dt} e^t = e^t$$ 7. So, $$\mathbf{r}'(t) = \langle 2t, \cos(t), e^t \rangle$$ 8. Example 2: Let $$\mathbf{r}(t) = \langle \ln(t), t^3, \cos(t) \rangle$$ for $t > 0$. 9. Differentiating each component: - $$\frac{d}{dt} \ln(t) = \frac{1}{t}$$ - $$\frac{d}{dt} t^3 = 3t^2$$ - $$\frac{d}{dt} \cos(t) = -\sin(t)$$ 10. So, $$\mathbf{r}'(t) = \left\langle \frac{1}{t}, 3t^2, -\sin(t) \right\rangle$$ This completes the definition and examples of derivatives of vector-valued functions.