Question

in this activity you will explore taking derivatives of vectors and related quantities. you will perform these operations throughout the course so this will be good practice. while some of the operations are straightforward, a few will be a bit more subtle. for each question think carefully about the order in which you should be doing things since this can dramatically change the result. consider the vector (vec{v} = 0.5t^8 hat{i} - 12t^2 hat{j}) using this vector, evaluate each of the following derivatives at (t = 1.6). derivative (left. \frac{dvec{v}}{dt}
ight|_{t=1.6} = square hat{i} + square hat{j}) magnitude of the derivative derivative of the magnitude magnitude of the integral integral of the magnitude

Explanation:

Step1: Differentiate the i - component

The i - component of \(\vec{v}\) is \(0.5t^{8}\). Using the power rule \(\frac{d}{dt}(t^{n})=nt^{n - 1}\), the derivative of \(0.5t^{8}\) with respect to \(t\) is \(0.5\times8t^{7}=4t^{7}\).

Step2: Evaluate the i - component derivative at \(t = 1.6\)

Substitute \(t = 1.6\) into \(4t^{7}\). First, calculate \(1.6^{7}\). \(1.6^{2}=2.56\), \(1.6^{3}=1.6\times2.56 = 4.096\), \(1.6^{4}=1.6\times4.096=6.5536\), \(1.6^{5}=1.6\times6.5536 = 10.48576\), \(1.6^{6}=1.6\times10.48576=16.777216\), \(1.6^{7}=1.6\times16.777216 = 26.8435456\). Then multiply by 4: \(4\times26.8435456 = 107.3741824\).

Step3: Differentiate the j - component

The j - component of \(\vec{v}\) is \(- 12t^{2}\). Using the power rule, the derivative with respect to \(t\) is \(-12\times2t=-24t\).

Step4: Evaluate the j - component derivative at \(t = 1.6\)

Substitute \(t = 1.6\) into \(-24t\): \(-24\times1.6=-38.4\).

Answer:

\(\frac{d\vec{v}}{dt}\big|_{t = 1.6}=107.3741824\hat{i}- 38.4\hat{j}\) (The first box should be filled with \(107.3741824\) and the second box with \(-38.4\))