61 - Calculus Questions Answers

y²= 4a(x+a)

so 2yy^' = 4a so a = yy^'/2

on putting a we get y²= 4yy^'/2(x+yy^'/2)      

so y² = yy^' (2x+yy^')  or y = 2xy^' + yy^'2    (1)

now on putting -1/y^' in the place of y^'

we get y² = -y/y^'[2x-y/y^']

so -yy^'2 = 2xy^' - y (2)

similarity of (1) and (2) shows that the given curve is self orthogonal 

limx-->0    (sin^-1x  - tan^-1x )/x³

use D L Hospital rule or expansion of sin^-1x and tan^-1x to solve it

two intersecting points means two solutions

multiply with the conjugate in numerator again and again you will get 

lim_x->∞ ((x+1)(x+2)(x+3)(x+4) - x⁴)/((x+1)(x+2)(x+3)(x+4))^1/4 + x )*((x+1)(x+2)(x+3)(x+4))^1/2 + x² )

after solving the numerator get a 3 power expression of x then take 3 power of x common, similarly take x and 2 power of x from the first and second denominator terms

its I. E. Maron 

limn->∞ 4ⁿ/ n!

= lim_n->∞ [4/1][4/2][4/3][4/4][4/5][4/6][4/7]................................[4/n]

besides first four bracket, all brackets are real numbers less than 1 so their product will be zero for n tends to infinite

since first one is a straight line so at every point only one tangent is possible so it is differentiable

but second one which is modulus has a sharp turn at x = -5/2 so two tangents at x = -5/2 are possible so is not differentiable.

perimeter = 4 so 2Y + X + πX/2 = 4

and area = XY + [π(X/2)²]/2

put Y from 1st equation to 2nd we get A = X[4-X{1+(π/2)}]/2 + πX²/8

now calculate dA/dX and then compare it to zero

X will be 8/(4+π)

this is a special type of differential equation in which p = dy/dx

its solution will be y = cx+(a/c)  here c is a constant

on applying p = dy/dx

(ydx-xdy)/dx = (xdx+ydy)/dx

or ydx-xdy = xdx+ydy

or dy/dx = (y-x)/(y+x)

now it is homogeneous