Jump to content

Related Rates question


Uker
 Share

Recommended Posts

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)

Link to comment
Share on other sites

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? :p:huh: or the √5 in general lol

appreciate it though man <33's

Link to comment
Share on other sites

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? :p:huh: 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..

Link to comment
Share on other sites

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? :huh: :huh: 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 :p

thanks - this can be closed

Link to comment
Share on other sites

Guest
This topic is now closed to further replies.
 Share

×
×
  • Create New...