1. **Given Problem:** Find the derivative of the vector function $$F(t) = \sin(t)\mathbf{i} + t^4\mathbf{j} - e^{2}\mathbf{k}$$ with respect to $t$. 2. **Derivative Calculation:** - The derivative of $\sin(t)$ with respect to $t$ is $\cos(t)$. - The derivative of $t^4$ with respect to $t$ is $4t^3$. - The derivative of the constant vector component $-e^{2}$ is $0$ because it does not depend on $t$. Therefore, $$\frac{dF}{dt} = \cos(t)\mathbf{i} + 4t^3\mathbf{j} + 0\mathbf{k} = \cos(t)\mathbf{i} + 4t^3\mathbf{j}$$ 3. **Finding the Magnitude $||F'(t)||$:** - The magnitude of a vector $\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ is $$||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2}$$ Here, $$a = \cos(t), \quad b = 4t^3, \quad c = 0$$ So, $$||F'(t)|| = \sqrt{\cos^2(t) + (4t^3)^2 + 0^2} = \sqrt{\cos^2(t) + 16t^6}$$ **Final answers:** $$\frac{dF}{dt} = \cos(t)\mathbf{i} + 4t^3\mathbf{j}$$ $$||F'(t)|| = \sqrt{\cos^2(t) + 16t^6}$$