find the rate of change of the distance between the origin and a moving point?

Question

on the graph of y=x^2+1 if dx/dt=2 centimeters per second

Answer 1

L = √(x² + y²)
ie L² = x² + y²
y = x² + 1
so L² = x² + (x² + 1)²
= x^4 + 3x² + 1
2LdL/dt = (4x³ + 6x)dx/dt
dL/dt = (4x³ + 6x)/L as dx/dt = 2
= 2x(2x² + 3)/√(x^4 + 3x² + 1) cm/s

Answer 2

s is the distance between a point on the graph and the origin.

s^2 = x^2 + y^2 = x^2 + (x^2 + 1)^2

2s ds/dt = 2x dx/dt + 2 (x^2 + 1) * 2x dx/dt

You need a specific point (x,y) to get a specific ds/dt, but you can plug dx/dt into the equation above to get something a bit more specific.

Answer 3

the distance r is given by:
r^2=x^2+y^2;
y=x^2+1;
so, r^2=x^2+(x^2+1)^2;
differentiating w.r.t time t we have:
2*r*dr/dt= (4*x^3+6*x)*dx/dt
hence dr/dt=(4*x^3+6*x)*(dx/dt)/(x^4+3*x^2+1)^0.5;
since dx/dt=2;
that gives dr/dt= (8*x^3+12*x)/(x^4+3*x^2+1)^0.5;
putting the value of x in the eqn the valocity of the point w.r.t origin can be found out.