A baseball diamond is a square 90 feet on a side. A player runs from first base to 2nd at a rate of 16 feet per second. So the distance between first base and 2nd pace is decreasing. So DXDT is going to be -16 feet per second. We want to know at what rate is the player's distance from third base changing when the player's 15 feet from first base. So looking at the picture they give us, we know that between 2nd and 30s ninety, we know that it's 15 feet between the first base and the player. So the X has to be 90 -, 15 or 75, and that's going to equal our y ^2. So when we solve our y ^2, we're going to get 15 square roots of 61. So if we solve for that, do our plugging and chugging, we get Y equal 15 ^2 to 61. So we know that y ^2 equal X ^2 + 90 ^2. Taking the derivative 2Y DYDT equal 2XD XDT, the derivative of that 90 ^2 is 0 because it was a constant. So DYDT is going to be X / y DXDT and at that specific location our X was 75 and our Y was 15 square roots of 61 and our DXDT was given as -16. So our DYDT should be -80 / sqrt 61. For Part B it says when the player is 15 feet from the first base, what are the rates of change of the angles? So tangent Theta two would equal X / 90. So the derivative is secant squared Theta 2D Theta 2DT equaling 190th DX DT. So D Theta two DT would equal cosine squared Theta 2 / 90 DX DT. Looking at the triangle, our cosine value is going to be the adjacent over the hypotenuse or 90 / 15 square roots of 61, and that's going to be squared times the 190th or divided by 90 times the DX DT of -16. So when we simplify this out, we get -332 / 305 radicals per radians per second. Now you could have used a different trig function if you wanted. If you had wanted to do secant of Theta two, you would have gotten y / 90, so secant Theta two, tangent Theta 2D, Theta D T would equal 190 DYDT. Taking the secant tangent to the other side, we would have had cosine Theta 2 and cotangent Theta 2 * 1 / 90 times the DYDT. Substituting those in we would have gotten the same answer. It wouldn't have mattered which direction which trig function we used, so we would have had -32 / 305 radians per second. Now we know that Theta one plus Theta 2 equal 90, so D Theta 1DT plus D Theta 2DT is got to equal 0 because once again the the derivative of a constant 0. Solving for D Theta one in terms of T, we get negative D Theta two DT. So we get +32 / 305 radians per second. Now the third one says as the player slides into second base, what are the rates of change of the angles Theta one and Theta two? So D Theta two DT is going to equal the cosine squared Theta 2 / 90 DX DT that was given earlier. And if we think about we're sliding into second, our X value is 0 or our Theta is actually zero. We don't have an angle anymore. So the cosine of 0 is one and 1 ^2 is 1. So 1 / 90 times the DX DT that was given was negative 16. So it's going to be -840 fifth radians per second and the D Theta one DT is going to be 845th radians per second. The answer in course compass, at least for my problem, is incorrect because they used 18 feet per second and it's not 18 feet per second. In my particular problem, it was 16 feet per second. So there is definitely an error in course compass on this particular problem on this example that I've done.