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Find the coordinates of the point P on the curve
y=64/(x^2) for x>0

where the segment of the tangent line at P that is cut off by the coordinate axes has its shortest length.

2006-11-08 14:48:37 · 2 answers · asked by thesekeys 3 in Science & Mathematics Mathematics

2 answers

Let P be (a,b). The slope of the tangent at P is found from the derivative of the curve: -128 / x³, giving the equation of the tangent as

          y - b = ( -128 / x³ ) ( x - a )

Now find where this tangent lines cuts the X and Y axes. Let x=0 and solve for y. Let y=0 and solve for x to get

          ( 0 , b + 128 / a² ) and ( a + a³b / 128 , 0 )

(a, b) is on the curve so b = 64 / a². Use this to simplify the two intercepts to

          ( 0 , 192 / a² ) and ( 3/2 a , 0 )

Use Pythagoras to find the length between these points as the square root of

          f(a) = 192² / a^4 + 9 / 4 a²

Now the problem is to minimize this expression (ignoring the square root, because it is an increasing function). This is a standard minimization calculus problem. Set f '(a) = 0, and check f "(a) > 0 to find that the minimum occurs when a = 4√2, so the required point is

          ( 4√2 , 2 ).

2006-11-08 17:24:16 · answer #1 · answered by p_ne_np 3 · 0 0

P: f c17969ba49117264eb77e9f66daacfdo17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfde17969ba49117264eb77e9f66daacfd y 17969ba49117264eb77e9f66daacfdx17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd whe17969ba49117264eb77e9f66daacfd x=0, 17969ba49117264eb77e9f66daacfdo y=3 Q: f c17969ba49117264eb77e9f66daacfdo17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd x 17969ba49117264eb77e9f66daacfdx17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd whe17969ba49117264eb77e9f66daacfd y =0, 17969ba49117264eb77e9f66daacfdo x = p17969ba49117264eb77e9f66daacfd/2 [17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd] radians) the 17969ba49117264eb77e9f66daacfdt17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfdght l17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfde PQ 17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd y= 3-(3/(p17969ba49117264eb77e9f66daacfd/2))x = 3-(6/p17969ba49117264eb77e9f66daacfd)x b) f'= -317969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfdx : 17969ba49117264eb77e9f66daacfdt Q f'=-3 y = -3(x-p17969ba49117264eb77e9f66daacfd/2) you c17969ba49117264eb77e9f66daacfd17969ba49117264eb77e9f66daacfd 17969ba49117264eb77e9f66daacfdo 17969ba49117264eb77e9f66daacfde17969ba49117264eb77e9f66daacfdt

2016-10-15 13:35:35 · answer #2 · answered by ? 4 · 0 0

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