Note: You need to have your calculator in radians (RAD) mode for this task (since the angles are given in radians - i.e., π is featured). a) A current waveform in a robotic arm may be described by... i_s=13 cos (2πft-(π/4))(A), where frequency, f=1Hz and t represents time. Make time (t) the subject of this formula and hence determine a point in time when the current waveform has a magnitude of +10A .

To make time \( t \) the subject of the formula, we start with the equation: \[ \mathrm{i}_{\mathrm{s}}=13 \cos \left(2 \pi \mathrm{ft}-\frac{\pi}{4}\right) \] Setting \( \mathrm{i}_{\mathrm{s}} = 10 \), we have: \[ 10 = 13 \cos \left(2 \pi (1)t - \frac{\pi}{4}\right) \] Now, dividing both sides by 13 gives us: \[ \frac{10}{13} = \cos \left(2 \pi t - \frac{\pi}{4}\right) \] Next, we take the inverse cosine (arccos) of both sides: \[ 2 \pi t - \frac{\pi}{4} = \cos^{-1}\left(\frac{10}{13}\right) \] Solving for \( t \): \[ 2 \pi t = \cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4} \] \[ t = \frac{1}{2\pi} \left(\cos^{-1}\left(\frac{10}{13}\right) + \frac{\pi}{4}\right) \] Calculating \( \cos^{-1}\left(\frac{10}{13}\right) \): Using a calculator, we find: \[ \cos^{-1}\left(\frac{10}{13}\right) \approx 0.4901 \, \text{radians} \] So we can substitute this value back in: \[ t \approx \frac{1}{2\pi} \left(0.4901 + 0.7854\right) \] \[ t \approx \frac{1}{2\pi} \cdot 1.2755 \approx 0.2034 \, \text{seconds} \] Thus, a point in time when the current waveform has a magnitude of +10 A is approximately \( t \approx 0.2034 \, \text{seconds} \).