Question

Given t(theta)=sin(theta)-2cos(theta) on the interval [0,2pi].

(a) Determine where the function is increasing and decreasing.


Note: Express the answer using interval notation. If the answer includes more than one interval write the intervals separated by the "union" symbol, U.

b) Determine where the function is concave up and concave down. Note: Express the answer using interval notation. If the answer includes more than one interval write the intervals separated by the "union" symbol, U

Answer 1

can you help me

Answer 2

plzz

Answer 3

No, I'm trying to figure out my own question.

Answer 4

\[\large t(\theta)=\sin \theta-2\cos \theta\]

A function can change from increasing to decreasing at points which we call `critical points`. So our first step is to find critical points of this function.
To do so we find the first derivative, and set it equal to zero, and solve for x.

\(\large t'(\theta)=0\)

Have you tried taking the derivative yet?

Answer 5

The derivative of that is 2sin(θ)+cos(θ)

Answer 6

So we need to solve for theta? Mmm ok lemme think real quick.

Answer 7

I know the interval is important, but I don't know how to use it.

Answer 8

This function is periodic, it will repeat its pattern over and over.
So we could find an infinite number of critical points.

The interval is just telling us, don't worry about the repeating pattern, just deal with it between 0 and 2pi.

Answer 9

If we set our first derivative equal to zero, we have,\[\large 2\sin \theta+\cos \theta=0\]
Let's subtract cosine from each side,\[\large 2\sin \theta=-\cos \theta\]
Dividing each side by -cosine gives,\[\large -2\frac{\sin \theta}{\cos \theta}=1\]
Then we'll divide each side by -2,\[\large \frac{\sin \theta}{\cos \theta}=-\frac{1}{2}\]
Recalling our trig identity, we can write this as,\[\large \tan \theta=-\frac{1}{2}\]

Understand up to this point? It's a bit of work to solve for theta :)

Answer 10

I think so

Answer 11

To solve for theta we can apply the arctangent function to both sides,\[\large \arctan\left[\tan \theta\right]=\arctan\left(-\frac{1}{2}\right)\]On the left, the arctangent and tangent are inverses of one another, so they'll "undo" one another.

\[\large \theta=\arctan\left(-\frac{1}{2}\right)\]
Hmm this is a weird problem, we don't have a nice reference angle for this :(

Answer 12

@satellite73 @phi