The temperature at a point (x, y) is t(x, y), measured in degrees celsius. a bug crawls so that its position after t seconds is given by x = 2 + t , y = 1 + 1 7 t, where x and y are measured in centimeters. the temperature function satisfies tx(3, 2) = 8 and ty(3, 2) = 7. how fast is the temperature rising on the bug's path after 7 seconds? (round your answer to two decimal places.)

373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km