Answer:
D(x) /dt = - 21,33 ft/s
Step-by-step explanation:
We have an equation, in which both x and y coodinates are function of t, then we take derivatives on both sides of the equation to obtain
3*x² + 8*y² = 11
6* x * D(x) /dt + 16* y * D(y)/dt = 0
6* x * D(x) /dt = - 16* y * D(y)/dt (1)
Now from problem statement we know:
D(y) /dt = 8 ft/s
And we are loking for D(x)/dt = ?? when particle passes the point ( 1,1)
x = 1 y = 1
Plugging these values in equaton (1)
6* x * D(x) /dt = - 16* y * D(y)/dt
6* D(x) /dt = - 16* 8
D(x) /dt = - 128 /6 ⇒ D(x) /dt = - 21,33 ft/s