Question

velocity and speed problem!

Answer 1

The max value of
\[A \cos \theta + B \sin \theta= \sqrt{A^2+B^2}\]
minimum value is
\[-\sqrt{A^2+B^2}\]

Answer 2

there have so much derivative point at v'(t)

Answer 3

Could you try now @akl3644 ?

Answer 4

i just graphy it

Answer 5

it's valid for both
\[A\cos \theta \pm B\sin \theta\]

Answer 6

i think it is not necessary to use tri-sub.?

Answer 7

just deriviate and check it graph would be better?

Answer 8

there have lots of v'(t)=0.

Answer 9

sorry. I read it wrong. Wait for sometime. I'll try to solve it

Answer 10

ok!

Answer 11

\[|x|=\sqrt{x^2}\]

Answer 12

\[\sqrt{(3\cos(4t)-4\sin(3t))^2}=s\]\[\sqrt{9\cos^2(4t)+16\sin^2(3t)-24\cos(4t)\sin(3t)}=s\]

Answer 13

We are trying to find minima/maxima here, so differentiate...

Answer 14

\[\frac{ds}{dt}=a=\frac{1}{2\sqrt{(9\cos^2(4t)+16\sin^2(3t)-24\cos(4y)\sin(3t)}}\frac{d(9\cos^2(4t)+16\sin^2(3t)-24\cos(4y)\sin(3t))}{dt}\]

Answer 15

\[\frac{d(9\cos^2(4t)+16\sin^2(3t)-24\cos(4y)\sin(3t))}{dt}=\]
\[9(2\cos(4t)*-4\sin(4t))+16(2\sin(3t)*3\cos(3t))-24((\cos(4t)*3\cos(3t))+(\sin(3t)*-4\sin(4t)))\]The end bit was\[-24((\cos(4t)*3\cos(3t))+(\sin(3t)*-4\sin(4t))) \] (I may have slipped up, check me.

-Set a=0
-Find t (at maxima/minima)
-find da/dt,
If at you t, da/dt= +ve, it is a minimum, if -ve, a maximum