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Example
Created on 2024-06-20Asked by Benjamin Allen (Solvelet student)
Find the derivative of the vector function r(t)=⟨t2,t3,sin⁡(t)⟩\mathbf{r}(t) = \langle t^2, t^3, \sin(t) \rangle.

Solution

To find the derivative of the vector function r(t)=⟨t2,t3,sin⁡(t)⟩\mathbf{r}(t) = \langle t^2, t^3, \sin(t) \rangle: 1. **Differentiate each component:** drdt=⟨ddt(t2),ddt(t3),ddt(sin⁡(t))⟩. \frac{d\mathbf{r}}{dt} = \left\langle \frac{d}{dt}(t^2), \frac{d}{dt}(t^3), \frac{d}{dt}(\sin(t)) \right\rangle. 2. **Compute the derivatives:** ddt(t2)=2t,ddt(t3)=3t2,ddt(sin⁡(t))=cos⁡(t). \frac{d}{dt}(t^2) = 2t, \quad \frac{d}{dt}(t^3) = 3t^2, \quad \frac{d}{dt}(\sin(t)) = \cos(t). 3. **Combine the results:** drdt=⟨2t,3t2,cos⁡(t)⟩. \frac{d\mathbf{r}}{dt} = \langle 2t, 3t^2, \cos(t) \rangle. 4. **Result:** The derivative of the vector function r(t)=⟨t2,t3,sin⁡(t)⟩\mathbf{r}(t) = \langle t^2, t^3, \sin(t) \rangle is: r′(t)=⟨2t,3t2,cos⁡(t)⟩. \mathbf{r}'(t) = \langle 2t, 3t^2, \cos(t) \rangle. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Michael Brown on Solvelet
1. Find the velocity and acceleration vectors of the particle whose position is given by the vector function r(t)=⟨cos⁡(t),sin⁡(t),t⟩\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle.2. Determine the equation of the tangent line to the curve described by the vector function r(t)=⟨t,t2,t3⟩ \mathbf{r}(t) = \langle t, t^2, t^3 \rangle at the point where t=2 t = 2 .,
Definition vector functions (receive some number of variables, and give back a vector) of form \(\mathbb{R}\mapsto \mathbb{R}^k\). They are frequently used to represent curves or surfaces in space. Ex: r(t)=⟨t,t2,t3⟩ (a curve in three dimensions)
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