
#1
Nov1611, 03:43 PM

P: 17

The problem is:
For the baseball diamond in exercise 33, suppose the player is running from first to second at a speed of 28 feet per second. Find the rate at which the distance from home plate is changing when the player is 30 feet from second base. (the picture basically shows a square with 90 ft sides) __________________________________________________ My solution Given: dy/dt = 28 ft/sec Find: ds/dt when x=30 90^{2}y^{2}=s^{2} 0 +2y(dy/dt)=2s(ds/dt) 2(60)(28)=2(30√13)(ds/dt) 56/(√13) ft/sec=ds/dt __________________________ I'm not really sure if its the correct answer because I always seem to mix up negative signs and I wasnt sure if this required one. If you really need to see the picture its found here: http://books.google.com/books?id=E5V...20base&f=false Page 151 #34 



#2
Nov1611, 08:32 PM

Mentor
P: 21,069

90^{2} + y^{2}=s^{2} 


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