The position of a particle moving in a straight line is given by s(t)=e^12t cos (5t) for t 0 where t is in seconds. If the particle changes direction at time T seconds, then T must satisfy the equation:

Question

The position of a particle moving in a straight line is given by s(t)=e^12t cos (5t)  for t > 0 where t is in seconds. If the particle changes direction at time T seconds, then T must satisfy the equation:

nathanjhw · Answer

tan(5T ) = 2.4

       tan(5T ) = 4.8

       cos(5T ) = 0

nathanjhw · Answer

attachment

nathanjhw · Answer

@dan815

nathanjhw · Answer

Yes

nathanjhw · Answer

I'm still confused though.

nathanjhw · Answer

@dan815

anonymous · Answer

Changing direction would imply that the derivative is 0 or undefined.  Basically T has to be a critical point.

nathanjhw · Answer

So then the answer would be cos(5T)=0?

nathanjhw · Answer

@wio

anonymous · Answer

$$
s'(t) = 12e^{12t}\cos(5t) - 5e^{12t}\sin(5t)  = e^{12t}\left(\frac{12}{5} -\tan(5t)\right)
$$If we want $s'(t)=0$, considering that $e^{12t}>0$ for all $t$, then we want $$
\frac{12}{5} -\tan(5t)=0 \implies \tan(5t) = \frac{12}{5} = 2.4 
$$

nathanjhw · Answer

Thank you very much @wio !

anonymous · Answer

Not sure if you learned anything

anonymous · Answer

I made a mistake anyway

anonymous · Answer

$$
s'(t) = 12e^{12t}\cos(5t) - 5e^{12t}\sin(5t)  = 5e^{12t}\cos(5t)\left(\frac{12}{5} -\tan(5t)\right)
$$

anonymous · Answer

So there are two potential equations to satisfy.

dan815 · Answer

Oops ignore what i write >_> i am tired.. i thought changing direction meant going all the way back