A function f is called homogeneous of degree n if it satisfies the equation f(tx, ty) = tnf(x, y) for all t, where n is a positive integer and f has continuous second-order partial derivatives. If f is homogeneous of degree n, show that fx(tx, ty) = tn − 1fx(x, y).

The answer is given below:

Step-by-step explanation:

Let u= tx,  v= ty Then

      d/dt [(f (u,v)]  = nt^n-1 f (x,y)  

Chain rule formula gives:

     ∂f/∂u x du/dt + ∂f/∂v x dv/dt = nt^n-1 f (x,y)

Therefore, x ∂f/∂u + y ∂f/∂v = nt^n-1 f (x,y) eq. 1

Put t=1 in eq. 1 :

=> x∂f/∂x + y ∂f/∂u = nf (x,y)

hope this !

step-by-step explanation:

d = distance between moon and sun, which in this problem is the hypotenuse of a right triangle.  

the distance y is the side opposite to angle x in the right triangle.   use the definition of the sine function from basic trigonometry.

sin(x) = y/d ⇒

d = y/sin(x)