Can some one check my work? Related Rates


by nyr
Tags: check, rates, work
nyr
nyr is offline
#1
Nov16-11, 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

902y2=s2
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
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Mark44
Mark44 is offline
#2
Nov16-11, 08:32 PM
Mentor
P: 21,069
Quote Quote by nyr View Post
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

902y2=s2
It looks like you left out the + sign on the left side. The equation should be
902 + y2=s2
Quote Quote by nyr View Post
0 +2y(dy/dt)=2s(ds/dt)
2(60)(28)=2(30√13)(ds/dt)
56/(√13) ft/sec=ds/dt
Looks fine to me.
Quote Quote by nyr View Post
__________________________

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


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