# College Math Help

College math has been known to keep students up all night, poring over their books and remembering very little the next day. If this situation sounds a bit too familiar, then its time to look for college math help online. College level math help is one the most eagerly sought after sections on math help websites. The sites are focused on topics covered in college level math like algebra, calculus, applied math, and statistics to name a few. Students who were struggling to pass, now find them selves enjoying the subject and the challenges it throws out thanks to online college math help.

## Math Help for College Students

Online help with college math offers a one on one teaching experience which inspires confidence and fosters learning. Online college help gives students the advantage of finding teachers who match their learning style. Students will also find the discussion boards and forums interesting as they have questions asked by other students, discussions with tutors and tips on how to deal with exam anxiety.

## Solved Examples

Question 1: Prove that the envelope of a plane, the sum of the squares of whose intercepts on the coordinate axes is constant, is a surface x2/3 + y2/3 + z2/3 = Constant.
Solution:
Let the equation of the plane be

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c}$ = 1                                    ......................(1)

where a2 + b2 + c2= k2                                            ......................(2)

Eliminating 'c' the equation of the plane becomes

F = $\frac{x}{a} + \frac{y}{b} + \frac{z}{\sqrt{k^2 - a^2 - b^2}}$ - 1 = 0

Therefore $\frac{\partial F}{\partial a}$ = 0

=> $\frac{-x}{a^2}- \frac{1}{2}z(k^2 - a^2 - b^2)^{\frac{-3}{2}}$ (-2a)            .....................(3)

and $\frac{\partial F }{\partial b}$ = 0

=> $\frac{-y}{b^2}- \frac{1}{2}z(k^2 - a^2 - b^2)^{\frac{-3}{2}}$(-2b)            ......................(4)

By equation (3) and equation (4) we have

$\frac{x}{a^2} * \frac{b^2}{y} = \frac{a}{b}$

=> $\frac{x}{a^3}$ = $\frac{y}{b^3}$ = $\frac{z}{c^3}$

or $\frac{\frac{x}{a} + \frac{y}{b} + \frac{z}{C}}{a^2 + b^2 + c^2} = \frac{1}{k^2}$         ........................(5)

From equation (5) we have

a = k2/3 x1/3, b = k2/3 y1/3, c = k2/3 z1/3

Put the values of a, b, c in equation (1), the envelope of the plane is

x2/3 + y2/3 + z2/3 = k2/3  = Constant

Question 2: Evaluate $\Delta$(Sinx Cos3x)
Solution:
$\Delta$(Sinx Cos3x)

= $\Delta${$\frac{1}{2}$(2Sinx Cos3x)}

or $\Delta${$\frac{1}{2}$(2Cos3x Sinx)}

[2Cos A Sin B = Sin(A + B) - Sin(A - B)]

= $\Delta${$\frac{1}{2}$(Sin 4x - Sin 2x)}

= $\frac{1}{2}${$\Delta$ Sin4x - $\Delta$ Sin2x}

= $\frac{1}{2}${Sin4(x + h) - Sin4x} - {Sin2(x + h) - Sin 2x}

[Sin A - Sin B = 2Cos($\frac{A + B}{2}$) Sin($\frac{A - B}{2}$)]

= $\frac{1}{2}${2Cos(4x + 2h)Sin2h - 2Cos(2x + h)Sinh}

= Cos(4x + 2h)Sin2h - Cos(2x + h)Sinh

= Sinh{2Cos(4x + 2h)Cosh - Cos(2x + h)}

[Sin2A = 2SinA CosB]

=> $\Delta$(Sinx Cos3x) = Sinh{2Cos(4x + 2h)Cosh - Cos(2x + h)}

Question 3: Find the general solution of the equation

$\frac{d^2y}{dx^2}$ - 2$\frac{dy}{dx}$ - 3y = 0
Solution:
Given differential equation

$\frac{d^2y}{dx^2}$ - 2$\frac{dy}{dx}$ - 3y = 0

or D2 y - 2Dy - 3y = 0

or (D2 - 2D - 3)y = 0

or  D2 - 2 D - 3 = 0

Solve for D,

D2 - 2 D - 3 = 0

=> D2 - 3D + D - 3 = 0

=> D(D - 3) + (D - 3) = 0

=> (D + 1)(D - 3) = 0

either D + 1 = 0   or   D - 3 = 0

=> D = -1, 3

The solution of the differential equation is

y = C1 e-x + C2 e3x