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Uker 

Mar 18 2008, 03:14 PM
Post
#1


*Uker 
No idea where to start............ thanks for help 


lotd 

Mar 18 2008, 09:50 PM
Post
#2



Okies, let's see how well I remember related rates lol..
So first off, when the player is halfway on his way to first base, his distance from first base is 45 feet. You can make a triangle from player/first base/second base. So you get a triangle with lengths 45, 90, 45*root5 which i will call sides a,b,c respectively. You know the rate the player is running at is 24 ft/s. This value is da/dt Now you set up your related rates problem. a^2 + b^2 = c^2 Now you take the derivative. b is a constant 90 and doesn't change so it goes to 0. you are left with: 2a*da/dt = 2c*dc/dt You are looking for the rate at distance from second base is decreasing, so that is dc/dt. Plug in your variables 2*45*24=2*45*root5*dc/dt Solve for dc/dt and you get 24/root5 If you form a triangle with third base now, you notice it is the same. In this case the rate the player's distance from second base is decreasing is the same at which his distance from 3rd base is increasing. So you answer is: a) Decreasing at 24/root5 ft/s or 24/root5 ft/s Increasing at 24/root5 ft/s Hope this helps. (and that it is correct lmao) 


Uker 

Mar 18 2008, 10:04 PM
Post
#3


*Uker 
Okies, let's see how well I remember related rates lol.. So first off, when the player is halfway on his way to first base, his distance from first base is 45 feet. You can make a triangle from player/first base/second base. So you get a triangle with lengths 45, 90, 45*root5 which i will call sides a,b,c respectively. You know the rate the player is running at is 24 ft/s. This value is da/dt Now you set up your related rates problem. a^2 + b^2 = c^2 Now you take the derivative. b is a constant 90 and doesn't change so it goes to 0. you are left with: 2a*da/dt = 2c*dc/dt You are looking for the rate at distance from second base is decreasing, so that is dc/dt. Plug in your variables 2*45*24=2*45*root5*dc/dt Solve for dc/dt and you get 24/root5 If you form a triangle with third base now, you notice it is the same. In this case the rate the player's distance from second base is decreasing is the same at which his distance from 3rd base is increasing. So you answer is: a) Decreasing at 24/root5 ft/s or 24/root5 ft/s Increasing at 24/root5 ft/s Hope this helps. (and that it is correct lmao) thank youuuu, but is root5 square root of 5? √5?? if so.....where did you get 45*√5? or the √5 in general lol appreciate it though man <33's 


lotd 

Mar 18 2008, 10:30 PM
Post
#4



Okies, let's see how well I remember related rates lol.. So first off, when the player is halfway on his way to first base, his distance from first base is 45 feet. You can make a triangle from player/first base/second base. So you get a triangle with lengths 45, 90, 45*root5 which i will call sides a,b,c respectively. You know the rate the player is running at is 24 ft/s. This value is da/dt Now you set up your related rates problem. a^2 + b^2 = c^2 Now you take the derivative. b is a constant 90 and doesn't change so it goes to 0. you are left with: 2a*da/dt = 2c*dc/dt You are looking for the rate at distance from second base is decreasing, so that is dc/dt. Plug in your variables 2*45*24=2*45*root5*dc/dt Solve for dc/dt and you get 24/root5 If you form a triangle with third base now, you notice it is the same. In this case the rate the player's distance from second base is decreasing is the same at which his distance from 3rd base is increasing. So you answer is: a) Decreasing at 24/root5 ft/s or 24/root5 ft/s Increasing at 24/root5 ft/s Hope this helps. (and that it is correct lmao) thank youuuu, but is root5 square root of 5? √5?? if so.....where did you get 45*√5? or the √5 in general lol appreciate it though man <33's root5 = square root of five. You get it from the Pythagorean theorem. 45^2 + 90^2 = C^2 C=45*root5 root5 is just easier to type than square root of 5 lol.. 


Uker 

Mar 18 2008, 11:10 PM
Post
#5


*Uker 
Okies, let's see how well I remember related rates lol.. So first off, when the player is halfway on his way to first base, his distance from first base is 45 feet. You can make a triangle from player/first base/second base. So you get a triangle with lengths 45, 90, 45*root5 which i will call sides a,b,c respectively. You know the rate the player is running at is 24 ft/s. This value is da/dt Now you set up your related rates problem. a^2 + b^2 = c^2 Now you take the derivative. b is a constant 90 and doesn't change so it goes to 0. you are left with: 2a*da/dt = 2c*dc/dt You are looking for the rate at distance from second base is decreasing, so that is dc/dt. Plug in your variables 2*45*24=2*45*root5*dc/dt Solve for dc/dt and you get 24/root5 If you form a triangle with third base now, you notice it is the same. In this case the rate the player's distance from second base is decreasing is the same at which his distance from 3rd base is increasing. So you answer is: a) Decreasing at 24/root5 ft/s or 24/root5 ft/s Increasing at 24/root5 ft/s Hope this helps. (and that it is correct lmao) thank youuuu, but is root5 square root of 5? √5?? if so.....where did you get 45*√5? or the √5 in general lol appreciate it though man <33's root5 = square root of five. You get it from the Pythagorean theorem. 45^2 + 90^2 = C^2 C=45*root5 root5 is just easier to type than square root of 5 lol.. hahha aight thanks  this can be closed 


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