Question

Evaluate (if possible) the vector-valued function at each given value of \(t\). \(\mathbf{r}(t)=\frac{1}{2} t^{2} \mathbf{i}-(t-1) \mathbf{j}\) (a) \(\mathbf{r}(1)\) (b) \(\mathbf{r}(0)\) (c) \(\mathbf{r}(s+1)\) (d) \(\mathbf{r}(2+\Delta t)-\mathbf{r}(2)\)

Short answer

The answers are: (a) \(\frac{1}{2}\mathbf{i}\), (b) \(-\mathbf{j}\), (c) \((s+1)\mathbf{i}+s\mathbf{j}\), (d) \( Δt^{2}\mathbf{i} + Δt\mathbf{j}\)

Step-by-step solution

Evaluate at t=1

Substitute \(t=1\) into \(\mathbf{r}(t)=\frac{1}{2} t^{2} \mathbf{i}-(t-1) \mathbf{j}\). Thus \(\mathbf{r}(1)=\frac{1}{2}*1^{2} \mathbf{i}-(1-1) \mathbf{j} = \frac{1}{2}\mathbf{i}'

Evaluate at t=0

Substitute \(t=0\) into \(\mathbf{r}(t)=\frac{1}{2} t^{2} \mathbf{i}-(t-1) \mathbf{j}\). Thus \(\mathbf{r}(0)=\frac{1}{2}*0^{2} \mathbf{i}-(0-1) \mathbf{j} = -1\mathbf{j}'

Evaluate at t=s+1

Substitute \(t=s+1\) into \(\mathbf{r}(t)=\frac{1}{2} t^{2} \mathbf{i}-(t-1) \mathbf{j}\). Thus \(\mathbf{r}(s+1)=\frac{1}{2}*(s+1)^{2} \mathbf{i}-((s+1)-1) \mathbf{j} = (s+1)\mathbf{i}+s\mathbf{j}\'

Evaluate the expression

Substitute t= 2 + Δt and t=2 into \(\mathbf{r}(t)\) and then calculate the difference. Thus, \(\mathbf{r}(2+\Delta t)-\mathbf{r}(2) = [\frac{1}{2}*(2+\Delta t)^{2} \mathbf{i}-((2+Δt)-1) \mathbf{j}]-[\frac{1}{2}*2^{2} \mathbf{i}-(2-1) \mathbf{j}]\) = \( Δt^{2}\mathbf{i} + Δt\mathbf{j}\)