63 - Calculus Questions Answers

let us consider the case of ellipse with x and y axes as their axes

eq. is x²/a² + y²/b²= 1

on differentiating we get 2x/a² + 2yy^'/b² = 0                                y'is first differential

so yy^'/x = -b²/a²

again differentiate and get answer. 

You should remember that you should differentiate as many times as the number of constants.

for ex. in the case of parabola only first diffrentiation is sufficient.

now complete it for all conics

x=rcosθ and y=rsinθ

so dx=-rsinθdθ  and dy = rcosθdθ

so p = -cotθ

so given eq. will become x-cotθy = (x-y)cosecθ 

on putting values of x and y we get 

0 = r(cosθ-sinθ)/sinθ

so tanθ = 1  so θ = nπ+π/4

so x = rcos(nπ+π/4) and y = rsin(nπ+π/4)

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)