Calculus question involving a dog.?
928648
Let Rover run to E 20 - x metres from R and then swim to P which is the hypotenuse of a right angled triangle with sides x and 5\n\nSo Rover runs (20 - x) m and swims √(x² + 25) m\n\nNow running rate = twice swimming rate.\nSo r = 2s\n\nSo time for running = (20 - x) /r and time for swimming = √(x² + 25)/s\n\nTotal time T = (20 - x) /2s + √(x² + 25)\n= 1/(2s)*(20 - x + 2√(x² + 25)\n\ndT/dx = 1/(2s)*( -1 + 2x/√(x² + 25)) = 0 for stationary points\nie -1 + 2x/√(x² + 25) = 0\nie  √(x² + 25) = 2x\nSquare both sides\nx² + 25 = 4x²\n3x² = 25\nx = ±5/√3\n= ±(5√3)/3 Since x > 0 the negative result is ignored.\nT = 1/2s (20 - (5√3)/3 + 2 √(100/3))\n= 1/2s (20 - (5√3)/3 + 20/√3)\n= 1/2s(20 + 5√3)\n\nIf he swam all the way T = √(425)/s = 5√(17)/s = 10√(17)/2s\nIf he walked to P and then swam straight in T = 20/2s + 5/s\n= 30/2s\n\nSo the quickest time is when he runs [20 - (5√3)/3] m down the beach and swims the rest of the way
619240
u can even use calculus to find this out
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
750604
As Stuart has said above, you need to use the product rule to work this out.  The product rule states:\n\nthe derivative of f(x)g(x) is f(x)g'(x) + f'(x)g(x)\n\nSo, when you differentiate sinx wrtx you get cos and when you differentiate 2x wrtx you get 2.  Plugging these into the formula above you get:\n\n(f(x)g(x))'=2x(cosx)+2(sinx)\n                =2xcosx + 2sinx\n\nHope that this is helpful!
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
711470
Try here:/\n\nhttp://archives.math.utk.edu/visual.calculus/
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
850369
The answer to this question is (b) 2/7.\n\nWhen removing two dogs there are four possible scenarios:\n\n(1) Remove poodle, then remove terrier.\n(2) Remove poodle, then remove another poodle.\n(3) Remove terrier, then remove another terrier.\n(4) Remove terrier, then remove poodle.\n\nIn each case, take the probability of the first even happening and multiply it by the probability of the second even happening. For instance:\n\nProbability that a poodle is first selected = 5(# of poodles originally) / 7(total # of dogs).\nProbability that a terrier is then selected = 2(# of terriers left) / 6(total # of dogs after a poodle is removed).\nMultiply 5/7 and 2/6 to get 10/42.\n\nRepeat this with each of the remaining three cases to get\n(1) 10/42\n(2) 20/42\n(3) 2/42\n(4) 10/42\n\n(Note that these add up to 1, which means that we have all of the possible cases)\n\nNow, just add the probabilities of the two cases with a terrier being selected second to get:\n\n10/42(first case) + 2/42(third case) = 12/42\n\nReduce 12/42 down to 2/7 to get your answer.\n\nI hope this helps, good luck with future probability problems!
572318
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
667073
Let x = number of boys and \ny = number of dogs, then\nx + y = 40 and\n2x + 4y = 100\nDivide the second equation by two and then subtract it from the first\nx + y = 40\nx + 2y = 50\n-y = -10\ny = 10\nx + y = 40\nx + 10 = 40\nx = 30\n30 boys, 10 dogs
887032
This is a fundamental theorem of calculus problem.\n\nThe derivative of the integral of t * sqrt (t^3 + 2t) dt from -1 to x is\n\nx * sqrt (x^3 + 2x)
838481
Well hey, you need help, ask the teacher! I know how to do this, with help from my nerdy Calculus BC friend, but it's really not that hard once you think about it...
908456
You're right that this is a parabola problem.  The specific parabola is:\n\nf(x) = x^2 + (20 - x)^2 = 2x^2 - 40x + 400\n\nNow, if you are taking calculus, just find the value where the first derivative is zero.  If you aren't in calculus yet, the minimum value is the bottom of the parabola.
3246
So for the first derivative:\n\n4(x-1)\n-----------\n3x^(2/3)\n\nThe second derivative:\n\n4(x+2)\n-----------\n9x^(5/3)
176854
You can graph it but probably that is not the answer your teacher is looking for. If you wanna do this by the good old method:\n\nf'(x)=3^x*ln(3)\n\nYou are looking for a point where the slope equals to 1 meaning 1 equals to your derivative.\n\n3^x*ln(3)=1\n\nNow use the calculator: x=-0.0856