How to prove the identity sin^2x + cos^2x = 1 ?
We can do this by many methods, a simple method is to take the derivative: f(x) = sin^2(x) + cos^2(x) Taking derivative: f '(x) = 2 sin(x) cos(x) + 2 cos(x) (-sin(x)) f '(x) = 2 sin(x) cos(x) - 2 cos(x) sin(x) f '(x) = 0 Since the derivative is zero everywhere the function must be a constant. Now to find that constant value: Take f(0) = sin^2(0) + cos^2(0) = 0 + 1 = 1 So, sin^2(x) + cos^2(x) = 1 everywhere. Hence proved.
If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, y(1 + xy) dx = x dy, then what is the value of f(-1/2) ?
Given curve is y = f(x) Given differential equation is : y(1 + xy) dx = x dy y/x (1+ xy) = dy/dx y = vx => y/x = v dy/dx = v + x dv/dx v(1+vx^2) = v + x dv/dx v^2x^2 = x dv/dx v^2x = dv/dx Integral of (x dx) = Integral of ( 1/v^2 dv) (x^2)/2 = (-1/v) +c (x^2)/2 = (-x/y) +c Put (1, -1) in the above equation: 1/2 = 1 + c => c = (-1/2) The equation become: (x^2)/2 = (-x/y) - 1/2 Now we have to find f(-1/2) Put x = (-1/2) ((-1/2)^2)/2 = -(-1/2)/y - (1/2) 1/8 = 1/2y - (1/2) y = 4/5 So, we got f(-1/2) as 4/5
A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h <<R). What is the minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field ?
Here, Earth's Radius = R Given Height = h Gravitational constant = G Let the Mass of Earth = M Let the Mass of Sattelite = m Now, Earth's Orbital velocity, v = (GM / R+h)^½ = (GM / R)^½ as h is very less than R Velocity required to escape is: (Due to Conservation of Energy) 1/2mv'^2 = (GMm/R+h) v' = (2GM / R+h)^½ = (2GM / R)^½ as h is very less than R Therefore, increase in velocity v' - v = (2GM / R)^½ - (GM / R)^½ = (2gR)^½ - (gR)^½ = (gR)^½ *(2^½ -1) So, the answer is (gR)^½ *(2^½ -1).