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Integral Calculus
5.find the curve for which the portion of tangent of the curve intercepted between the coordinate axis is of constant length.
6.find the family of curves which are such that the area of rectangle constructed on the absicca of any point and the initial ordinate of the tangent at that point is constant.
[also plz explain wht does initial ordinate of the tangent at that point mean...]
Comments (12)
let our curve be y = f(x)
eqn of tangent at (X,Y) is y - Y = f'(X) {x - X} where Y = f(X) and f'(X) is the slope of tangent at (X,Y)
y- intercept = Y - X f'(X)
x-intercept = X - Y/f'(X)
From now on , my notation is y' = f'(X) , y =Y , x=X just for the sake of simplicity in typing . Hope any1 of u wont mind.
so y intercept = y - xy' & x intercept = x - y/y'
A/Q, (x- intercept)2 + (y-intercept)2 = k (a constant)
(y - xy')2 + (x - y/y')2 = k
Since length is constant we can differentiate it and equate to 0
2(y - xy')(y' - y' - xy'') + 2(x - y/y')(1 - 1 + yy''/y'2) = 0
solving it we get x2y'4 - xyy'3 + xyy' - y2 = 0
xy'3(xy' - y) + y(xy' - y) = 0
(xy'3 + y)(xy' - y) = 0
either y'3 = -y/x or y' =y/x
If y' = y/x it gives ln(y) = ln(x) + c i.e. y = cx a straight line
If y' = - (y/x)1/3 y-1/3 dy + x-1/3 dx = 0
on integratin it gives y2/3/(2/3) + x2/3/(2/3) = a
x2/3 + y2/3 = 2a/3
let us choose a constant p such that p2/3 = 2a/3
this gives x2/3 + y2/3 = a2/3
this solution is accepted and the straght line solution is rejected as it is a straight line , it's tangent is the line itself .
So answer is x2/3 + y2/3 = a2/3
Note :- This is a hypocycloid .
Its parametric coordinates are (acos3t , asin3t).
Its the only curve for which the length of tangent at any point intercepted b/w the coordinate axis is constant .(as i have already shown)
You can remember the above mentioned points about the curve hypocycloid.
hypocycloid is generated by the trace of a fixed point on a small circle that rolls within a larger circle.
For more details about this curve , you may refer http://en.wikipedia.org/wiki/Hypocycloid
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as we saw in earlier question
assume y = f(x) is our curve
tangent at any point (x,y is Y - y = f'(x) (X - x)
y intercept = y - xy'
and abscissa is of course x
so the sides of rectangle are x & y - xy'
Area = x(y - xy') = k
y' + (-1/x)y = -k/x2
this is simple linear differential equation (the one which involves the use of integrating factor)
now please solve yourself and gain some moral back . I dont want to snatch it from you.
the answer is xy = k + cx2 Answer
hello dear
let the point be (h,k)
equation of tangent is::
(y-k)/(x-h) = dy/dx = m
( y-k ) = mx - mh
y intercept ::
(0,k-mh)
x intercept :
[(mh -k)/m ,0]
sum of intercept ::
(h - k/m) + (k -mh) = C
put m = dy/dx and h =x and k = y
(x - y /m) + (y - xm) = c
above is a differential equation,solve thi sand u will get the curve
if any problem in solving differential equation ,nudge me
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Bet we can't solve them? Why? Is it because you can't solve them yourself? lol