Well thanks for clarifying that up you guys.
I have one more similar question with which I need some help. I've gotten a bit further with this one but not enough:
I concluded that the limit equals 0 by trying a few paths (hopefully this time it was right):
\lim_{\{x,y\}\to \{1,0\}} \...
I updated my post above, to show that I did try some other paths too.
So if both limits are different, then it would mean that the limit does not exist.
I guess that WA isn't perfect after all. :)
Hi,
If we are approaching from the path x = y^3, then
lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3
= lim (x,y) -> (0,0) of (y)*y^2 / y^3+y^3 =
= lim(x,y) -> (0,0) of y^3 / 2y^3. = 1/2
That seems to make sense.
However, if we use the path y=x:
lim (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3
= lim...
Homework Statement
Evaluate or show that the limit DNE.
Limit as (x,y) -> (0,0) of (x^1/3)*y^2 / x+y^3.
The Attempt at a Solution
I tried approaching from multiple paths, and it seems that the limit is equal to 0. I used the delta-epsilon method to prove the limit but I've been stuck so far...
That was extremely thorough Mathstatnoob, thank you for taking the time to post the answer. I appreciate it very much. Through your steps, I've learned how to manipulate little pieces here and there to arrive at the needed answer. :)
If anyone could confirm my Q1 and Q5 solutions, I'd be...
Hello everyone, I am in need of a little assistance. I have a homework assignment due soon that consists of 5 questions. Of which, I have done all but the 4th one. I did start on it but I'm not sure where to go from there. I also would like for someone to confirm my solutions for question 1 and...
In that case:
Ax+By+Cz+D=0
(5)(2)+(-4)(-1)+7(8)+D=0
10+4+56+D=0
D=-70
Therefore, the Cart. Eq is 5x-4y+7z-70=0. Right?
And any luck on the 2nd question?
That makes sense.
Perhaps, they are the same? Just to be sure in the equation [x,y,z] = (-1,-2,-3) + s[5,-4,7], [5,-4,7] is the direction vector right?
If that is the case, could we use Ax+By+Cz+D=0 and plug in 5 for A; -4 for B; 7 for C? And x,y,z would be 2, -1 and 8 respectively to solve...
Um I THINK they are parallel? Something in my head pops up about the normal being perpendicular to the line. So if the line is perpendicular to the plane, they are parallel?
I'm likely wrong.
Homework Statement
Write the Cartesian equation for the plane containing the point (2,-1,8) and perpendicular to the line [x,y,z] = [1,-2,-3] + s[5,-4,7].
The Attempt at a Solution
The situation is that I have my Calc. + Vectors exam tomorrow morning and I'm just going through some...
You did a lot of unnecessary stuff there. Start on the left side. Cotangent (-x) is negative in the 4th quadrant, so what can that be expressed as? Cosine is positive in 4th quadrant so what can that be expressed as? Sine is negative in 4th quadrant, once again, adjust the sin(-x) into something...
In your solution, you did not even mention pi/4. The double angle is as follows:
2cos^{2}x-1=0
cos2x=0
2x=pi/2
x=pi/4
That is only one solution, and you are still missing it. Check where you went wrong.
Mspike, don't make that mistake. The cosx don't divide out. Since you have already pretty much attained your answer, this is what should have happened:
\frac{2+2cosx}{sinx+sinxcosx}
=\frac{2(1+cosx)}{sinx(1+cosx)} The (1+cosx) in the numerator and denominator divide out.
=\frac{2}{sinx}
=2cscx
Your final compounded value is incorrect. The answer is about $10475.76.
Now that you have the compounded value, you can see that the interest you earn in that one year is $10475.76-7700=$2775.76. I will leave it up to you to try the formula FV = PV (1+i)^n to find the answer that I got...
Turn the denominator into \frac{cos(x)+1}{cos(x)}.
So that would look like \frac{\frac{sin^{2}(x)}{cos(x)}}{\frac{cos(x)+1}{cos(x)}}.
What icystrike showed is the faster way but since you're already this far, you can try what I suggested.
Sorry, I didn't mean .34 pi. I just meant .34 rad.
And as for the sin(30) = 2, likely no. That's why I said to him to not base his work on my steps. I was hoping someone would correct it and I'd learn something new. I just recall something about cot x = a. tan x = (1/a) x = tan- (1/a). I guess...
You factored incorrectly. That should be (3sinx-1)(sinx-2)=0
So sin x = 1/3 and 2
Take the sine inverse of the two.
Sin x = 2. csc 30 = 2 so sin- (1/2) = x. x=30 degrees or pi/6 rad.
sin x = 1/3. sin- (1/3)=x. x=19.47 degrees or .34pi.
That's what I would've done but I'm not sure if that...
Nevermind, I thought of it quite a bit and I think I've got the concept.
But just to make sure:
First 100 km/h is converted into 2777.78 cm/s. That is the speed at which the tire travels. So basically, every second, it moves 2777.78 cm. This would make it the arc length.
Now we have to look...
Homework Statement
"A car is traveling at 100 km/h and the tire of the car has a radius of 36cm. Find the number of revolutions per second."
The Attempt at a Solution
100 km/h * (10,000,000 cm/km) * (1h/3600 secs) = 2777.77777778 cm/s is the speed of the car.
Θ = a/r
Θ = (2777.78) / (36)
Θ...
Here are the two special triangles I've used in the past: http://fouss.pbworks.com/f/special%20triangle%203.JPG and http://fouss.pbworks.com/f/special%20triangle%202.JPG and recall that sin (a-b)=(sin a)(cos b)-(sin b)(cos a)
Now sin \frac{-pi}{12} = sin (\frac{pi}{6} - \frac{pi}{4}) = sin...
Nope, no mistakes. I'm in Pre-Calc at the moment too, none of this stuff so far. We worked on this kind of stuff in grade 11 if memory serves me right.