Question

A Baseball Diamond is a square 90ft on a side. A player runs from first base to second base at a rate of 16 ft/sec.

-At what rate is the player's distance from third base changing when the player is 30ft from first base?

Answer 1

goy R 16 telec Güven 904 30 +84 OT I 9=30 Nhu 4th base a) To find using pythagoras theorem in AABL formed by ghel & god base AC² = A B² + BC² x² = (90)² + (60-y)" Differentiating both side wont. & andy - 0 + -2(90-4) dy - 0 out when y=30 n = √(30)² + 90 -30)² = 50) + (60² x= 108-167 m) substituting values of nly in equ. 2)(x) ly = -2(90- y) 216.334 (de) - 120 dy (Giren d = 16HHA) I 216.334 dx = (120) (16) du = -8.875 ft/sec

90 9o y 6) To find do, and do in AABC, we know tono, BC = AB tano, = AB = goy Using eq. tone) and diffentiating both side dt a tand) - I go by) sec) do, Boyd (90-4) = (BU E1) (90-y)-dy. dt dt got a dy now, sec(0.) = sony substituting values of uy and dy ni do = go n=108.167 16H/se 30.123 and we know in AABE 6 + 9, -50 : 100 0, +, = 90 , : 90-6, differentiating both sides do, do, at at