Question

What is the y-value of the global minimum for y=3e^(−t)*cos(t)? Help please!

Answer 1

Have you considered the 1st derivative? It might give you some insight. On the other hand, it could be very subtle.

Answer 2

I did take the derivative and got -3e^(-t)(cost+sint) = 0 but I don't know where to go from there.

Answer 3

Okay, where does that expression take on the value zero (0)?

Answer 4

I just set it equal to zero because there was an example in the book that did that. The book said the minimum is when dy/dt equals zero.

Answer 5

Interestingly, in this case, that does not actually provide a GLOBAL minimum. See if you can figure out why.

It may help to know that \(\sin(x) + \cos(x) = \sqrt{2}\cdot\cos\left(\dfrac{\pi}{4} - x\right)\)

Answer 6

So I should set the inside of the function to that? Sorry I still very confused.

Answer 7

*I'm

Answer 8

We just need to know where it is zero.

\(\cos\left(\dfrac{\pi}{4} - x\right) = 0\;at\;\dfrac{\pi}{4} - x = \dfrac{\pi}{2} + k\pi\), where k is any integer. Solving for x, we get \(x = -\dfrac{\pi}{4} + k\pi\), again, where k is an integer.

What does that tell us? It's zero infinitely many times!!!

Answer 9

So the minimum is 0?

Answer 10

No, this is where the 1st Derivative is zero. This is where the original function has AT LEAST a LOCAL Minimum or Maximum. Trouble is, one problem, anyway, is that there are infinitely many. Can we learn anything from this? Can we find a GLOBAL minimum?

Answer 11

Hmmm can we find one when t is bigger than or equal to zero? I think we can.

Answer 12

Why do we want to know that? Minima and maxima are where the 1st derivative is zero (0). Only zero (0) matters for this.

Answer 13

Without calculus we may be able to see how the original function behaves.
y=3e^(−t)*cos(t)
cos(t) is confined to -1 to + 1
For t > 0, e^(−t) becomes smaller and smaller approaching zero as t gets larger and larger
For t < 0, e^(−t) becomes larger and larger approaching infinity as t gets more and more negative.
So on the positive side, the curve approaches 0.
But on the negative side the curve keeps oscillating with increasing amplitude.
So the global minimum for y is negative infinity and global maximum is positive infinity.

Answer 14

The global minimum is NOT negative infinity, since that isn't actually anything. There is no global minimum. The local minima increase in the negative direction without bound as t increases in the negative direction.

Answer 15

Yes. The correct answer would be no global minimum. But the curve does approach +/- infinity as t approaches -infinity.

Answer 16

I tend to consider such language incorrect. In the Real Numbers, there is no such value as "infinity". Thus, it is not appropriate to approach "infinity". The proper language is to say that it increases without bound.

Not everyone agrees with me.

Answer 17

Calculus books are replete with references to phrases such as "t tends to infinity" or in a series "n approaching infinity", etc.
But hopefully mandy's question has been answered here.

Answer 18

Indeed.

Answer 19

I attached a picture of the example we're supposed to go off of from the book and they were able to find a minimum value for y on t greater than or equal to zero.

Answer 20

Like an exact value not infinity. The point I get thrown off at is when they turn the sin/cos into tangent.

Answer 21

You didn't say \(t \ge 0\). This is very important.

Way back up, we pick k = 1 and produce the least positive value of t such that the derivative is zero, \(t = \dfrac{3\pi}{4}\)

Answer 22

So I can then plug in 3pi/4 into the equation but I thought it should be a different number from the original equation with was e^(-t)*cos(t) and this one has a three in the front: 3e^(-t)*cos(t) doesn't that make a difference?

Answer 23

The value of t doesn't care if there is a 3 in front. However, you need the ORIGINAL equation to solve the problem.

Answer 24

So you're saying I should times the answer of the original equation by 3? Because I did that and that answer still wasn't right.

Answer 25

"i should times the answer" -- Use the work "multiply".

No, I'm saying you need the ORIGINAL equation. Why would you multiply it by 3 again?

Answer 26

Because the equation I'm working with is 3e^(-t)*cos(t). I'm sorry I'm really confused, I don't understand what you are trying to explain.

Answer 27

Nevermind I got the answer, I still had no clue what you were talking about but I got the right answer it's -.201 I just used a graph.

Answer 28

\(3\cdot e^{-3\pi/4}\cdot\cos(3\pi/4) = -0.201059219\)