Posted on 16 December 2016 - 12:24 AM
if 3 (yx^2) = y then what does dy/dx equal?ah that one was too easy.
what is the slope of the line tangent to the curve y = arcsin at x = 1/2?
Posted on 16 December 2016 - 12:58 AM
Zaverus wrote
Just you.
Posted on 16 December 2016 - 01:00 AM
silverteeth wrote
Solve:
Root 3 Root x+1 = Root 3x-5
(The root x+1 is under the root that contains the 3)
Hm, I'm not quite sure if I'm reading this correctly, but I'll do what I see.
sqrt(3(x+1)) = sqrt(3x-5)
3x+3 = 3x-5.
No solution.
Posted on 16 December 2016 - 01:04 AM
Limeintine wrote
I think what you're after is point-slope form.
Given any slope m and a point (x1,y1), you can define a unique line through that point.
Explicitly, m(x-x1) = y-y1, and this comes directly from choosing an arbitrary point (x,y) and substituting it into the slope formula:
m = (y-y1)/(x-x1).
Last edited on 16 December 2016 - 03:42 AM by AlbertEinstein
iHackLiTE wrote
ah that one was too easy.
what is the slope of the line tangent to the curve y = arcsin at x = 1/2?
1.) Use implicit differentiation.
(d/dx) (3yx^2) = (d/dx)y.
3(dy/dx)x^2 + 6yx = (dy/dx).
6yx = (dy/dx)(1-3x^2).
(dy/dx) = (6yx)/(1-3x^2).
One could also put this in terms of solely x by solving for y in 3yx^2 = y —> y = (3x^2-1)^(-1), giving
(dy/dx) = (-6x)/(1-3x^2)^2.
Or you could have just solved for y from the start and differentiated explicitly in terms of x.
2.) If I recall correctly, (d/dx)(arcsinx) = (1-x^2)^(-1/2).
If that's the case, then the slope of the tangent line at x = 1/2 is simply the derivative evaluated at x = 1/2.
Then, m = (3/4)^(-1/2) = (4/3)^(1/2) = 2/sqrt(3) (= (2*sqrt(3))/3, for you "rationalize the denominator" folk).
Edit: After a little thinking, I remembered the way to derive the inverse trig derviatives.
Let's do y = arccos(x). Then, cos(y) = x. Implicitly differentiate with respect to x to obtain -sin(y) (dy/dx) = 1.
Thus, dy/dx = -csc(y). Use right triangle trignometry to find csc(y) given than cos(y) = x/1 = adjacent/hypotenuse. Thus, csc(y) = (1-x^2)^(-1/2).
Hence, dy/dx = -(1-x^2)^(-1/2).
Posted on 16 December 2016 - 01:39 AM
Automatically DeletedLast edited on 16 December 2016 - 07:05 AM by ThatOneCombo
x^3 + y^3 + z^3 = 29x^3 + y^3 + z^3 = 30
Last edited on 16 December 2016 - 07:14 AM by AlbertEinstein
ThatOneCombo wrote
x^3 + y^3 + z^3 = 30
I'm not seeing that there could be any solution to this system.
If (x1,y1,z1) were a solution to the first equation, then how could it also possibly be the same as that number plus one?
We'd be looking for numbers that satisfy a = a+1, and those simply don't exist.
Last edited on 16 December 2016 - 07:18 AM by ThatOneCombo
AlbertEinstein wrote
ThatOneCombo wrote...
I'm not seeing that there could be any solution to this system.
If (x1,y1,z1) were a solution to the first equation, then how could it also possibly be the same as that number plus one?
We'd be looking for numbers that satisfy a = a+1, and those simply don't exist.
That was a challenge math thing that was posted in the hallways, you can use only use integers and the two equations can have different values for the variables.
Posted on 16 December 2016 - 07:28 AM
ThatOneCombo wrote
AlbertEinstein wrote...
That was a challenge math thing that was posted in the hallways, you can use only use integers and the two equations can have different values for the variables.
First one's straightforward: 1^3+1^3+3^3 = 1+1+27 = 29.
Second one, haha: https://ckrao.wordpress.com/2012/04/10/integers-equal-to-the-sum-of-three-cubes/.
Code a program to go through millions and millions of numbers to get that one, or just look at the link.
Last edited on 16 December 2016 - 09:14 AM by Jinxful
What is the smallest thing? I can think of a quarkPosted on 16 December 2016 - 09:13 AM
Automatically DeletedPosted on 16 December 2016 - 10:00 AM
I have a headachePosted on 16 December 2016 - 10:34 AM
Automatically DeletedLast edited on 16 December 2016 - 01:28 PM by silverteeth
AlbertEinstein wrote
silverteeth wrote...
Hm, I'm not quite sure if I'm reading this correctly, but I'll do what I see.
sqrt(3(x+1)) = sqrt(3x-5)
3x+3 = 3x-5.
No solution.
13 + root 105 all over 6
Posted on 16 December 2016 - 01:54 PM
Jinxful wrote
Posted on 16 December 2016 - 01:56 PM
silverteeth wrote
AlbertEinstein wrote...
13 + root 105 all over 6
1.) Stop using "root" to talk about square roots. There are "cube roots" and "sixteenth roots" as well.
2.) Use parentheses, because only you know what you meant by your equation.
Posted on 16 December 2016 - 01:58 PM
Jinxful wrote
The "Planck length" and strings are the things that come to mind.
Posted on 16 December 2016 - 02:37 PM
Would you be interested in answer philosophy questions?