Question

Find \((a) r^{\prime \prime}(t)\) and \((b) r^{\prime}(t) \cdot r^{\prime \prime}(t)\). $$ \mathbf{r}(t)=\left\langle e^{-t}, t^{2}, \tan t\right\rangle $$

Short answer

(a) \( r''(t) = \left\langle e^{-t}, 2, 2\sec(t)\tan(t) \right\rangle \). (b) \( r'(t) \cdot r''(t) = -(e^{-2t}) + 4t + 2\sec^2(t) \tan(t) \)

Step-by-step solution

Calculate the first derivative \( r'(t) \)

The derivative of \( r(t) \) with respect to \( t \) is calculated by taking the derivatives of its components functions. The derivative of \( e^{-t} \) is \( -e^{-t} \), the derivative of \( t^2 \) is \( 2t \), and the derivative of \( \tan t \) is \( \sec^2(t) \). Therefore, \( r'(t) = \left\langle -e^{-t}, 2t, \sec^2(t) \right\rangle \)

Calculate the second derivative \( r''(t) \)

The second derivative of \( r(t) \) with respect to \( t \) is calculated by taking the derivatives of the component functions in \( r'(t) \). The derivative of \( -e^{-t} \) is \( e^{-t} \), the derivative of \( 2t \) is \( 2 \), and the derivative of \( \sec^2(t) \) is \( 2\sec(t)\tan(t) \). Therefore, \( r''(t) = \left\langle e^{-t}, 2, 2\sec(t)\tan(t) \right\rangle \)

Calculate the dot product \( r'(t) \cdot r''(t) \)

The dot product of two vectors is computed by multiplying the corresponding components and adding the resulting products. So, \( r'(t) \cdot r''(t) = -e^{-t}\cdot e^{-t} + 2t\cdot 2 + \sec^2(t) \cdot 2\sec(t)\tan(t) = -(e^{-2t}) + 4t + 2\sec^2(t) \tan(t) \).