The math help thread

[COLOR="Green"]I got this type of problem in calc today. I know the formula to use, but can't figure out what to do.

y=X^3, X[SIZE="1"]0[/SIZE]=3
Find the instantaneous velocity of the object at X[SIZE="1"]0[/SIZE].

The formula is:

lim...y={f(X) - f(X[SIZE="1"]0[/SIZE])} / (X - X[SIZE="1"]0[/SIZE])
X->X[SIZE="1"]0[/SIZE]

The numbers and equation for y are just placeholders, because they were on a test and I don't have it back yet. What exactly am I supposed to be doing here? :f6:[/COLOR]

If you haven't learned differentiation, it literally represents the 'rate' at which a function changes.

Say you have y = x^3 to represent an object's motion, position, as measured in x and y axes. The differentiation (or derivative, whatever) of that function will give a function which describes the "rate" at which that motion is changing, in other words, its "speed" (because movement/time = speed).
Applying the same understanding again, the differentiation of a speed is the object's acceleration (or how much speed it changes in x time).

As for the problem itself, I haven't done Math in years and even when I did, I wasn't really focused but I guess I can give it a try.

lim y = [f(x) - f(xo)]/(x - xo)
x->xo

Suggests the "limitation" that the function reaches as x approaches xo, or simply understood as the value of y at xo. They prefer to use "limitation" since more than often there isn't a specific value for y at x = xo.

lim y = [f(x) - f(xo)]/(x - xo) =
x->xo
lim [(x^3 - 27)/(x - 3)] =
x->3
lim (x^2 + 3x + 9) = 27.
x->3

I have a feeling I'm wrong on many levels but I'd like to use this opportunity to know where I'm wrong lol.

That's suppose to be 27, no? I mean, 9 + 9 + 9 != 27 for some reason?

/reversenitpick

Oh my. People actually use that? .
Thank you for derivative shortcuts.

Russt: I did the same thing with the 3's. Sheesh.

That's suppose to be 27, no? I mean, 9 + 9 + 9 != 27 for some reason?

/reversenitpick

Right. I checked it with Power Rule and got 27, then looked at the expression again and somehow missed for the second time that 3 * 3 is not 6.

@ Levi
Typically the way it's taught (at my school at least) is that they introduce the definition of the derivative, with the clunky limit stuff, and make you do a couple homework assignments using that. Then they teach you Power Rule and the rest.

Of course, half the students find out about the shortcuts before they're supposed to.

Oh my. People actually use that? .
Thank you for derivative shortcuts.

Believe me, that is NOT a derivative shortcut! It spells pain everywhere when you get to the complicated crap like 3x*sin(sqrt(5x^2 + 6)).

Either that or I just suck at it o.O

Believe me, that is NOT a derivative shortcut! It spells pain everywhere when you get to the complicated crap like 3x*sin(sqrt(5x^2 + 6)).

Either that or I just suck at it o.O

Complicated or just long?

Is there any way to get 50 mi/h into ft / s^2? "A car is traveling 50 mi/h, decelerating at 22 ft / s ^2, blah blah..." I figure I need to make them the same units but the s^2 is kind of throwing me off. Wouldn't 4400 ft/s work? ):

@ Russt: Yeah, my professor made us learn it like that and use it like that on our first test. Excruciatingly painful given the alternate route...

Oh my. People actually use that? .
Thank you for derivative shortcuts.

Russt: I did the same thing with the 3's. Sheesh.

Now you know why I'm getting a graphing calculator for Christmas. But the teacher put up "YOU MUST SHOW ALL WORK OR YOU GET NO CREDIT, NO DERIVATIVES."

Complicated or just long?

Is there any way to get 50 mi/h into ft / s^2? "A car is traveling 50 mi/h, decelerating at 22 ft / s ^2, blah blah..." I figure I need to make them the same units but the s^2 is kind of throwing me off. Wouldn't 4400 ft/s work? ):

@ Russt: Yeah, my professor made us learn it like that and use it like that on our first test. Excruciatingly painful given the alternate route...

Units, son! Use the power of labeling numbers with units!

Code:

[U]50 mi[/U] * [U]5280 ft[/U] * [U]  hr   [/U]
hr        mi     3600 s

Now you have the velocity on the same units as the acceleration.

Oh and no you can't change ft/s into ft/s^2. The acceleration value describes how much the velocity changes per second or w/e the denominator unit is.

Nitpick.

Hey man, Hamiltonian Quaternions deserve some love, too.

Why must I suck at Physics?

The ball launcher in a pinball machine has a spring that has a force constant of 1.2 N/cm. The surface on which the ball moves is inclined 10.0 degrees with the horizontal. If the sprint is initially compressed 5.00 cm, find the launching speed of a 100g ball when the plunger is released. Friction and the mass of the plunger are negligible.

So, I was assuming I am to apply conservation of energy; that gets me an answer close to what the back of the book says, but it's an entire .1 m/s off. :\

Another question might be on its way. D:

Did you include both the potential energy in the spring, and the potential energy by moving the ball 5 cm up a 10 degree slope? I think the second would be small enough to account for that kind of error.

Silly me. I had the PE on the wrong side.

Wspring = KE + PE ? I originally had Wspring + PE = KE. When I changed it, it worked out. :]

dy/dx=cos x * e^{y + sin x}

I'm suppose to find y. Told the answer's y= -ln( C - sin x ), but no idea why.
How does one arrive at this answer?

[COLOR="Magenta"]Help me pass calculus. I'm failing and panicking.

We're on related rates, optimization, and describing graphs without the calculator.

I'm dying.


Here's a problem I just can't figure out. I've got to find the following
a) derivative (straight forward, and easy)
b) minimum points
c) maximum points
d) increasing
e) decreasing
f) second derivative (again straight forward)
g) concave up
h) concave down
i) points of inflection

using the formula f(x) = ln ((x^2) +1)


please don't just solve it, EXPLAIN it for me >: I suck at this class.

Thank you![/COLOR]

a) derivative
you said it's straightforward and easy so I'll just give it: 2x/(x^2 + 1)
b) minimum points
This has to be when the derivative is 0, so:
2x/(x^2 + 1) = 0
2x = 0
x = 0
And the derivative changes from negative to positive at this point, which is how we know it's a minimum.
c) maximum points
Same as (b) except the derivative must change from positive to negative. There are none for this function (it goes to infinity)
d) It's increasing when the derivative > 0.
2x/(x^2 + 1) > 0
2x > 0
x > 0
e) Decreasing when derivative < 0
2x/(x^2 + 1) < 0
2x < 0
x < 0
f) second derivative is a peach but it's -2(x^2-1)/(x^2+1)^2

I gotta run so really quickly...
g) concave up when d^2 > 0
h) concave down when d^2 < 0
i) inflection points when d^2 changes sign

I can try to help if what you're looking for is explanation.

[COLOR="Magenta"]Here's a problem I just can't figure out. I've got to find the following
a) derivative (straight forward, and easy)[/COLOR]
the rate of a function changing (f(x) changes as x changes)

[COLOR="Magenta"]b) minimum points
c) maximum points[/COLOR]
These 2 points are the spots where your graph changes direction. To put more simply: once something hits its top, it starts to decline and vice-versa. Imagine: going up, hitting A (maximum point #1), starts to go down, hits B (minimum point #1), starts to go back up, goes OVER A and may or may not hit C (maximum point #2.)
That's why A, B and C are called local maximum/minimum, they're the highest (or lowest) since on both sides the function graph is beneath (or above them.) These are points where the graph "bends"
So if you look at this, you'll see that when the graph "bends", the tangent line literally lies horizontally, suggesting that the first derivative is 0.

[COLOR="Magenta"]d) increasing
e) decreasing[/COLOR]
if x > xo and f(x) > f(xo) or if x < xo and f(x) < xo => increasing
otherwise, decreasing.

I think this part compares the way your function changes in comparison to the way your variable changes. If your x is increasing (x > xo) but your graph is going down (y < yo)
then the function is "growing" in the opposite direction of your variable's growth. Not really sure here.

f) second derivative (again straight forward)
Describes the changes of the tangent line (which is ... changing)

I'm not so sure about this myself lol.

[COLOR="Magenta"]g) concave up
h) concave down[/COLOR]
In the same gif image above, you see how the tangent line changes from "spinning" clockwise to counter-clockwise. The direction in which it "spins" determines its concavity. To make long stories short: mountain = concave down, valley = concave up. Why? The algebraic value of your tangent line continuously decreases on the concave down curve (hence why it's -down-) and vice-versa.

i) points of inflection
No idea. I only learned that this is where the graph changes from being concave up to concave down.

Tried my almost-best. Hope I helped.

If you have a function f in an interval [a,b], then the minimum and maximum points are the points where the derivate either is 0 or undefined, or a or b.. For (a,b), this doesn't count (and thus doesn't count for infinity and such).

The function is increasing whenever the derivative is positive, and decreasing whenever the derivative is negative. That is, in an interval (a,b) given that a and b is either 0, undefined or (negative) infinity.

Concave up and down is the same as above, just for the double-derivative.

Points of inflection is where the double-derivative is 0 or undefined.

Noah

"Jack has a job that pays $15,000 for the 1st year with a raise of $1,000 at the end of each year thereafter. Jill has a job that pays $7500 for the first six months with a raise of $250 at the end of every 6 months thereafter. Who has earned the greater total income after 10 years and by how much? Write a formula for each income using sigma notation and then evaluate."

I actually got this problem, and it turns out that Jill has more of the total income by $2,500. The answer matches the one in the book.

But...it doesn't make sense. Jack is starting with more money, plus, his raise is higher at the end of every year...since Jill gets a raise of $500 per year, right? Plus, Jill starts out with less money, so I don't get how Jill's total income is higher than Jack's.

15000 + 1000t = Jack
7500 + 500t = Jill
t=10

15000 + 1000(10) = 25000 (Jack)
7500 + 500(10)= 12500 (Jill)


Am I doing something wrong...?

15000 + 1000t = Jack
7500 + 500t = Jill
t=10

15000 + 1000(10) = 25000 (Jack)
7500 + 500(10)= 12500 (Jill)


Am I doing something wrong...?

This seems like the "logical" way...and Jill has $1225 in your example by the way, since she doesn't get the raise on the first half of the first year.

But I used the sigma notation way, and it produced a different answer...which matches the one in the book, but it isn't logical...