so i have a few of these, i think i have got the first few ok,

but the next two are now confusing as they are a little more complex, at least to me since this is a new topic, never heard of it before

i = sin (5t^3 +6t -3)


i = 4e^cos(t)

then i have to find the max and minima of
y= x3 - 3x + 2

 

so i have a few of these, i think i have got the first few ok,

but the next two are now confusing as they are a little more complex, at least to me since this is a new topic, never heard of it before

i = sin (5t^3 +6t -3)


i = 4e^cos(t)

then i have to find the max and minima of
y= x3 - 3x + 2

Google is your friend:
Look for "Chain rule for differentiation"

 

thanks for the tip, i will do, i think its Di/Dt

 

Thanks for the tip, I will do, I think its Di/Dt

Tell me what you get, and I'll tell you if it is correct.

 

(cos 5t^3 +6t - 3) (15^2t +6 ) ??

 

(cos 5t^3 +6t - 3) (15t^2 +6 ) <-- correction

Yes, except you did not write the result correctly. You wrote 15^2t, but I knew what you meant.

\( i\;=\;sin(5t^3+6t+3) \)
\( \cfrac{di}{dt}\;=\;cos(5t^3+6t+3)(15t^2+6) \)

 

so i have a few of these, i think i have got the first few ok,

but the next two are now confusing as they are a little more complex, at least to me since this is a new topic, never heard of it before

i = sin (5t^3 +6t -3)


i = 4e^cos(t)

then i have to find the max and minima of
y= x3 - 3x + 2

Hello there,

You say you have two sinusoidal functions then you say you have to find the min and max of the purely algebraic function.
So do you have to find the min and max of the sinusoidal functions too? If so, how do you have to express the result?
I ask because the first one with the sin() in it is more complicated than the other two. It's likely to have a host of min's and max's.

 

He just has to differential the two sinusoidal expressions. Separate from that, he has a third equation that he needs to find the min and max of.

The tricky part of the second one is what type of minima and maxima is he supposed to find? Local? Global? Over some interval?

 

He just has to differential the two sinusoidal expressions. Separate from that, he has a third equation that he needs to find the min and max of.

The tricky part of the second one is what type of minima and maxima is he supposed to find? Local? Global? Over some interval?

Hi,

You said he just has to differentiate the two sinusoidals, but then say finding the min and max of the second one is tricky, but if he doesnt have to do the second one for min and max how is that tricky? The second one is one of the sinusoidals. Wondering what you meant.

 

Read the initial post again more carefully. It is actually pretty clearly laid out.

so i have a few of these, i think i have got the first few ok,

but the next two are now confusing as they are a little more complex, at least to me since this is a new topic, never heard of it before

i = sin (5t^3 +6t -3)


i = 4e^cos(t)

This is his first task -- to differentiate (as indicated by the thread title) those two expressions. Nothing about minima and maxima here.

then i have to find the max and minima of
y= x3 - 3x + 2

This is his second task -- to find the minima and maxima of that one equation (almost certainly meant to be x^3 and x3).

The tricky part of the second one (i.e., the second part, or task) is what type of minima and maxima is he supposed to find.

He is NOT being asked to find the minima or maxima of EITHER sinusoidal expression. Those are part of his FIRST task.

 

Read the initial post again more carefully. It is actually pretty clearly laid out.


This is his first task -- to differentiate (as indicated by the thread title) those two expressions. Nothing about minima and maxima here.



This is his second task -- to find the minima and maxima of that one equation (almost certainly meant to be x^3 and x3).

The tricky part of the second one (i.e., the second part, or task) is what type of minima and maxima is he supposed to find.

He is NOT being asked to find the minima or maxima of EITHER sinusoidal expression. Those are part of his FIRST task.

Hi,

Oh i see, so when you said second part you did not mean the second equation you meant the second grouping. No problem.

As to the tricky part, i don't see how he could be hunting for anything but the relative min and/or max unless you want to take the infinities as a result too but can that be said to be a min or max. Also, since differentiation was the first part (first grouping) then it makes sense that it would be calling for the relative min and max. Could be wrong but that's what it looks like to me.

 

That's my guess as well, but engineering isn't about guessing. Either the problem should make it clear was was being sought (since, hopefully, the distinctions have been covered), or the context of the assignment should have strongly implied which is being sought (and that's very possibly the case, here). Unfortunately, we lack that context, so we risk leading him down the wrong track if we assume too much about what that context was -- and it's hard not to do that and we all do it from time to time.

 

That's my guess as well, but engineering isn't about guessing. Either the problem should make it clear was was being sought (since, hopefully, the distinctions have been covered), or the context of the assignment should have strongly implied which is being sought (and that's very possibly the case, here). Unfortunately, we lack that context, so we risk leading him down the wrong track if we assume too much about what that context was -- and it's hard not to do that and we all do it from time to time.

Yes i have to agree.

I base my guess on the fact that they were given two preceding exercises that were to find the derivatives, which imply they were leading up to the next exercise which is the natural progression of teaching and when applying that rule we find two finite results, and also that infinity being a max although not finite is a kind of open question, combined with the lack of a well defined interval for the functions.

 

sorry for confusion, i may have not written them out completely clearly

Taks 1
Diferentiate the following function, then use t to represent time = 3 seconds

a. i = sin (5t^3 +6t -3)

b. i = 4e^cos(t)

c. i = 2e (^3t+4)

Task 2
Locate the co-ordinates of the turning points for the following function and then decide if it is a maxima or minima

y= 7x^2 - 3x

my answer so far
dy/dx = 0 d(7x^2 - 3x) / dx = 0 d(7x^2)^dx / dx - d(3x)/dx = 0 2 x 7 x X^2-1 - 3 x Xv^1-1 = 0
14x - 3 = 0 x = 14/3 ?? at this point i dont know how to confirm max or min

Task 3
find the maxima and minima of the function y= x^2 - 3x + 2

 

Task 2
Locate the co-ordinates of the turning points for the following function and then decide if it is a maxima or minima

y= 7x^2 - 3x

my answer so far
dy/dx = 0 d(7x^2 - 3x) / dx = 0 d(7x^2)^dx / dx - d(3x)/dx = 0 2 x 7 x X^2-1 - 3 x Xv^1-1 = 0
14x - 3 = 0 x = 14/3 ?? at this point i dont know how to confirm max or min

Need to be more careful with your notation -- it WILL trip you up from time to time if you don't.

d(7x^2)^dx / dx
should be
d(7x^2)/dx

2 x 7 x X^2-1 - 3 x Xv^1-1 = 0

Don't use 'x' for multiplication when 'x' is a variable. 'X' and 'x' are not the same variable, so don't swap between them to avoid confusion with 'x' for multiplication. Instead, use '*' or '·' for multiplication.

Remember order of operations -- exponentiation is done before addition/subtraction -- so
X^2-1
is
(X^2) - 1
and not
X^(2-1)

Be careful not to introduce new things.
3 x Xv^1-1
What the heck is 'v'?

Instead of
2 x 7 x X^2-1 - 3 x Xv^1-1 = 0
Use
2*7x^(2-1) - 3x^(1-1) = 0

As for how to tell if it is a min or a max, consider the behavior of dy/dx in both cases.

If y is a max, that means that it was increasing before the peak and then decreasing after it. If a function is increasing, is its slope positive or negative?

What does this mean for dy/dx before and after the peak?

What does this mean for the slope of dy/dx before, at, and after the peak?

How can you determine whether the slope of dy/dx is positive or negative at the peak?

 

sorry about these errors, one is a typo and i using x and X is confusing so i will avoid doing this again.
Let me take another look at this

 

sorry for confusion, i may have not written them out completely clearly

Taks 1
Diferentiate the following function, then use t to represent time = 3 seconds

a. i = sin (5t^3 +6t -3)

b. i = 4e^cos(t)

c. i = 2e (^3t+4)

Task 2
Locate the co-ordinates of the turning points for the following function and then decide if it is a maxima or minima

y= 7x^2 - 3x

my answer so far
dy/dx = 0 d(7x^2 - 3x) / dx = 0 d(7x^2)^dx / dx - d(3x)/dx = 0 2 x 7 x X^2-1 - 3 x Xv^1-1 = 0
14x - 3 = 0 x = 14/3 ?? at this point i dont know how to confirm max or min

Task 3
find the maxima and minima of the function y= x^2 - 3x + 2

Wow, your notation and other stuff is really, really bad, you have to pay more attention to this or you will not be able to communicate your meanings. I'll point out a few things too.

First of course is the notation.
You should use the asterisk for multiplication as that has been accepted as such since 1970 maybe before that, and most math software also uses that. I also have to suggest that you always use a sign for multiplication because if you dont you will even confuse yourself sometimes.
So dont type y=3x instead type y=3*x. Dont be lazy with this when you are asking questions.
You also have to watch out for your exponentiation notation. There is no such sequence "(^" that i know of in math.

Your separation of expressions:
Dont type y=x+1 k=3 u^3=8
instead type either:
y=x+1, k=3, u^3=8
or use separate lines for each expression.

Last, your basic division for some reason.
14*x-3=0
does not come out to x=14/3. Check that.

 

so Task 2 question should be
\[ y = 7x^2 - 3x \]

\[ x = 3/14 \]

14 > 0 so the function will be minima ?
why do the equations centre like that on the post ?

 

What's your reasoning for basing your decision on 14 being greater than 0?

 

What's your reasoning for basing your decision on 14 being greater than 0?

I can guess a test for concavity, which suggests either a min or a max.
(Sign of the second derivative)