In the xy-plane, a particle moves along the parabola y=x2−x with a constant speed of 210−−√ units per second. If dxdt>0, what is the value of dydt when the particle is at the point (2, 2) ?

  3√21 units/second

Step-by-step explanation:

Let s', x', and y' represent the derivatives of the respective variable with respect to time. We are given s' = √210 units/second, and we want to find y'.

We also have the relation ...

  (s')^2 = (y')^2 +(x')^2 . . . . . from the Pythagorean theorem

We can differentiate the given equation for y to get ...

  dy/dx = 2x -1 = 2(2) -1 = 3 . . . . . at x=2

We also know ...

  y' = (dy/dx)(x') = 3x'

  x' = y'/3

Using this in the above equation for s', we get ...

  (√210)^2 = (y')^2 + (y'/3)^2 = (10/9)(y')^2

  (9/10)(210) = (y')^2

  3√21 = y'

The value of dy/dt is 3√21 units per second at the point (2, 2).

d.) 48 units

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.