On the 5th November 1854 a smaller British and (yes) French army beat off an assault by the Russians at Inkerman in the Crimea. It was known as “The Soldier’s Battle” as men fought small engagements due to poor visibility in dense fog.
The Russians had massed 32000 men on the Allied flank and headed for the 2700 man 2nd Division, commanded today by the aggressive Pennefather. Instead of falling back in the face of superior numbers, he advanced. The British had their rifles to thank this day as they took a terrible toll on the musket armed Russian Infantry, who were hemmed in by the valley’s bottle neck shape. The British 2nd Division pushed the Russians back onto their reinforcements and should have been routed by the Russians’ numbers, but the fog and the British Light Division saved them. Three successive Russian commanders were killed in this engagement.
The Russians other 15000 men approached and assailed the Sandbag Battery, but they were routed by 300 British defenders vaulting the wall, blunting the lead Battalions, who were then attacked in the flank. More Russian attacks ensured the Battery exchanged hands several times.
The British 4th Division was not as lucky. Arriving on the field, its flanking move was itself flanked and its commander, Cathcart, killed. This enabled the Russians to advance, but not for long. They were soon driven off by French units arriving from their camps and made no more headway.
The battle was lost and they had to withdraw.
This was the last time the Russians tried to defeat the Allied troops in the field. Despite this reverse, however, the Russian attack had seriously stalled the Allies from capturing Sevastopol. They had to instead, spend one harsh winter on the heights overlooking the city, before it fell in September of 1855.
The British suffered 2573 casualties, the French 1800 and the Russians 11959.
More calculus help
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Consider a curve of x(y^2)-x^(3)y=6.
The derivative is: (3(x^2)y-y^2) / (2xy-x^3)
I need to find when the derivative does not exist (tangent line vertical).
So far, as we only need to see if the denominator is 0, I set 2xy-x^3=0.
Then, I factored this to get x(2y-x^2)=0.
I know from here that when x=0, there is a vertical tangent line. However, I think there is more than that. What should I do to find the rest, or if I am doing this wrong, how do I solve it?
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Curious.
Is that XYY-X^(3Y)?
(question is about the second term. Need to know if the three and the y are powers of X.)
Let me know tonight, and I’ll help ya with it tomorrow.
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I need this for tomorrow.
It is x to the third times y, not x to the 3y.
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Solve for y:
X(Y^2)-Y(X^3)=0
Add Y(X^3) to both sides of the equalityX(Y^2)=Y(X^3)
Divide both sides by YXY=X^3
Divide both sides by XY=X^2 is your solution.
The graph of Y=X^2 exists for all Y >= 0.
The graph of Y=X^2 exists for all X (-oo < X < oo) where oo = infinityThe derivative of Y=X^2 is:
Y’=2X
2X is a linear function and has no points of discontinuity.
Both f(x) and f’(x) are continuous functions.
BTW, you did not take the derivative correctly when you attempted to derive the tangent line off the original function. To do this you would have to use implicit differentiation.
For instance, X^3 * Y using implicit differentiation is not 3X^2 * Y, it is 3X^2 * Y * Y’ where Y’ represents DY/DX.
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It equals 6, not zero.
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