Question

Absolute min and max from f(t) = 2 cos t + sin 2t, [0, π/2]

Answer 1

Mmm ok, find a derivative yet?

Answer 2

graphing it may be an easy solution.

Answer 3

use desmos

https://www.desmos.com/calculator

Answer 4

-2sint+cost ?

Answer 5

Hmmmm not quite :O

Answer 6

:(

Answer 7

\[\Large\rm (\sin2t)'=(\cos2t)\color{royalblue}{(2t)'}\]\[\Large\rm (\sin2t)'=(\cos2t)\color{orangered}{(2)}\]Chain rule, yah? :o

Answer 8

why are we using chain rule ?

Answer 9

Wut? 0_o

Because ... we ... always... use... chain rule :o

Answer 10

oh sorry, I thought we always used it when there was a power ...Like (3x+5y)^2

Answer 11

Chain rule is always necessary when the `inner function` is more than just "x".
Woops, "t" in this case.

Example:\[\Large\rm \frac{d}{dt}(t)^2=2(t)\color{royalblue}{\frac{d}{dt}(t)}=2(t)\color{orangered}{(1)}\]In this example, notice that by applying the chain rule, we gain nothing new?
Yah it's not necessary here.
But if the inner function was more than just t,
Example:\[\Large\rm \frac{d}{dt}(3t)^2=2(3t)\color{royalblue}{\frac{d}{dt}(3t)}=2(3t)\color{orangered}{(3)}\]

Answer 12

In our problem:
Since our inner function was 2t instead of t, we needed to chain rule.

Answer 13

Chain rule is by far the hardest of the quick shortcut rules to master :(
Spend some time practicing it.

Answer 14

oh alright, Soooo then sin2t = cost(2) would 2cost be -2sint ?

Answer 15

The derivative of the first term looks good.

D 2cos(t) = -2sin(t).

We're still having some trouble with the other one though :(

Answer 16

Ignore the chain rule for just a sec,\[\Large\rm \frac{d}{dt}\sin(2t)=\cos(2t)\]Yes?
For some reason your 2t keeps turning into a t.

Answer 17

If we properly apply our chain rule:\[\Large\rm \frac{d}{dt}\sin(2t)=\cos(2t)(2t)'\]
\[\Large\rm \frac{d}{dt}\sin(2t)=2\cos(2t)\]Chain rule has no effect on our inner function.
It shouldn't change it.
We're making a copy of it, and multiplying by the derivative of that copy on the outside.

Answer 18

Oops sorry. Then it would be = -2sin(t) + 2cos(2t) ?

Answer 19

Mmm ok looks good,\[\Large\rm f'(t)=-2\sin (t)+2\cos(2t)\]

Answer 20

Looking for critical points:\[\Large\rm 0=-2\sin (t)+2\cos(2t)\]

Answer 21

It's going to be a little tricky from here.
Do you remember your `Cosine Double Angle Identity`?
It shows up in three different forms.

Answer 22

This is the one that we want:\[\Large\rm \cos(2t)=2\cos^2(t)-1\]

Answer 23

Oh yeah I remember that.

Answer 24

Woops my bad, we have a sin(t) in our problem.
I wrote the wrong one :(
That's not the one we want.

Answer 25

This is the one we want:
\[\Large\rm \color{orangered}{\cos(2t)=1-2\sin^2(t)}\]

Answer 26

aha thats what I wrote :)

Answer 27

\[\Large\rm 0=-2\sin (t)+2\color{orangered}{\cos(2t)}\]hehe

Answer 28

So that will give us a bunch of sin(t)'s which will make it easier to work with.

Answer 29

ohh okay !

Answer 30

\[\Large\rm 0=-2\sin (t)+2\color{orangered}{\left[1-2\sin^2(t)\right]}\]

Answer 31

let x = sin ?

Answer 32

Distribute some stuff...
Divide a -2 out of everything...
eventually we get to:\[\Large\rm 0=2sin^2(t)+\sin(t)-1\]A substitution? :)
Mmmm yah that's a good idea!

Answer 33

Leading to something like this, yah? :o
\[\Large\rm 0=2x^2+x-1\]

Answer 34

I skipped a few algebra steps in there >.<
I think you can handle those hehe

Answer 35

Solve for x using your quadratic formula, yah?

Answer 36

brb i gotta get some chocolate milk all up in mah belly

Answer 37

where you at lady? :3
figure it out?

Answer 38

Sorry I fell asleep ! Ill look at it right now

Answer 39

Got 1/2 and -1

Answer 40

\[2x^2+2x-x-1=0,2x \left( x+1 \right)-1\left( x+1 \right)=0,\left( x+1 \right)\left( 2x-1 \right)=0\]
x=-1,1/2 is correct

Answer 41

\[\sin t=-1=-\sin \frac{ \pi }{ 2 }=\sin \left( \pi+\frac{ \pi }{ 2 } \right)=\sin \frac{ 3 \pi }{ 2 }\]
\[t=\frac{ 3 \pi }{ 2 } does \not belong \to \left[ 0,\frac{ \pi }{ 2 } \right]\]
hence rejected.
\[\sin t=\frac{ 1 }{ 2 }=\sin \frac{ \pi }{ 6 },t=\frac{ \pi }{ 6 }\in \left[ 0,\frac{ \pi }{ 2 } \right]\]
find f(t) at t=0,pi/6,pi/2
and see which is max. and which is min.