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).

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2