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Ricky, that is correct but only because the coefficient of p is a negative one. I was just pointing out the actual steps for any type of integration of the form f(x)/g(x).
I'm sorry, I forgot about the F1 - F2 equation, but it really is no different than what I stated above. My last post was F1 + F2. The only difference in that calculation and the subtraction one is that you must use the negative components of F2. All of the summation and equation still hold true.
F = √[(3cos90° - 5cos170°)² + (3sin90° - 5sin170°)²]
F = √[24.24615776 + 4.54439691], arctan(2.131759112/4.924038765)
F ≈ 5.37N, 23.41°
You can break up as many vectors as you want to add or subtract and use the Pythagorean theorem.
∑F = √[(∑x)² + (∑y)²]
Specifically the resultant force is;
F = √[(∑x)² + (∑y)²], arctan(∑y/∑x)
I actually think that this method is easier because you can simply break all the forces into two component columns and add them up. Once you do that you can just plug them in to the above equation.
For the original equations posted this is what would happen.
F1 = 3N, 90° and F2 = 5N, 170°
F = √[(3cos90° + 5cos170°)² + (3sin90° + 5sin170°)²]
F = √[24.24615776 + 14.96328757], arctan(3.868240888/-4.924038765)
F ≈ 6.26N, -38.15°
I'm thinking that toru missed the shift key in problem 2, making it 5(m+n)-(3m+2n).
5m + 5n - 3m - 2n = 2m + 3n.
They are asking you to differentiate 1/(1-p) with respect to p.
With fractions differentiation works like this;
1. Take the derivative of the numerator and multiply it by the divisor.
2. Take the derivative of the divisor and multiply it by the negative of the numerator.
3. Add 1. and 2. together and divide this by the divisor squared.
or... the derivative of f(x)/g(x) = [f'(x)g(x) - f(x)g'(x)] / g(x)²
In you above equation; (by the steps above)
[0(1-p) - 1(-1)] / (1-p)² = 1/(1-p)²
V = ∫2πx(f(x)-g(x))dx
2π∫x - xlnx dx = π[x²]0,e - 2π[x²lnx / 2]1,e + 2π[x²/4]1,e
Note that I had to use integration by parts for g(x).
V = πe² - πe² + 2π(e²/4 - 1/4) = π/2 (e²-1) ≈ 10.0359u²
Also note the upper and lower limits for the integrations above. We had to exclude the values of x lower than 1 for the logarithmic part because that would have been negative and subtracted from the shaded region above the x axis. I hope that this clears up some of the questions above.
John, note that even though you attempted to use the disk method instead of the cylindrical shell method that the volume you tried to compute was for the region below the lnx function. If you were to use the disk method the integral above should be subtracted from the integral of y=1 function.
Ricky, thank you for correcting me. I was confusing the second derivative test for local extrema. (blush)
What defines an inflection point is that both the first and second derivatives equal zero. The velocity and the acceleration are zero at a turning point, specifically an inflection is the moment when the velocity or rate of change is reversing direction.
When working on the above question it is found that the function does not exist. Mikau correctly eliminated c and d in the beginning, but failed to use both derivatives together to eliminate a and b as well.
f(x) = ax³ + bx²
f'(x) = 3ax² + 2bx
f''(x) = 6ax + 2b
Because an inflection occurs at x=1;
3a + 2b = 0 and 6a + 2b = 0
3a + 2b = 6a + 2b, which only works if both a and b are zero.
y = 0 does not satisfy any of the conditions originally proposed.
This might sound silly, but did you try to take the derivative and set it to zero?
I'll take an ill-attempted stab.
I might be nitpicking but VR Hawks posted a solution of 400L when the answer was 400m³. That would be 400,000L of water.
What? Mathsyperson if you take a pencil and make points with until you cover the earth you will only have a little under 328 million marks to choose from. You can't figure H from that?
Not curious...the area formula for a cone is 1/3 the area of a cylinder. Actually a cone is just a tapered cylinder anyway.
John, if you take any polygon and draw lines from the vertices to the center you will get n equal triangles with bases the length s and a vertice angle of 360/n°.
Since the height of any given triangle is s/2tan(180/n) the area of any triangle there will be;
A = (s/2) [s/2tan(180/n)] = s²/(4tan[180/n])
So the total area is number of sides times the area of any of the triangles which gives the formula you have above. But there was no calculus needed.
I know that John is clever. He is always trying to work things out on his own. A great way of really understanding the logic behind many concepts. I will provide the proof over on the other thread because he seemed to want that there.
John, you do not need calculus for finding the area of polygons.
Area = ns² / (4tan[180/n]) for anypolygon with equal sides numbering greater than 3
n = number of sides
s = length of side
This is used with an assumption of degrees being used and not radians for the trigonomic function. I am sure that you can figure out how to convert this to use for radians.
God factored out x-3 from the original equation. (x-3)(x²+3x+3) = x³-6x-9
The volume of the cylinder is the base time the height and you are given that information.
V = 100× 12 = 1200
If the cylinder is 1/3 full then the water volume is 1/3 the volume of the cylinder;
1200/3 = 400 m^3
At distance x the angle to the top of the lighthouse is 49°.
So the H = x tanθ or x tan 49°
At distance (x + 110) meters the angle is 22° so,
H = (x + 110) tan 22°
The height does not change so...
x tan 22° + 110 tan 22° = x tan 49
Therefore x = 110 tan 22° / (tan 49° - tan 22°)
Since H = x tan 49° = (110 tan 22° tan 49°) / (tan 49° - tan 22°)
H ≈ 68.5 meters
Whenever you see a log a + log b it says the same thing as log (ab). And whenever you see log a - log b it is the same as log (a/b).
So log 64 - log 128 = log (64/128) and log 32 + log 256 = log (256×32)
Now you have [log (64/128)] + [log (256×32)]
This then reduces to log [(64/128)(256×32)]
log 4096 ≈ 3.61236