Distance from home plate

Modigliani

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Dec 20, 2008
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Player running between 2nd base and 3rd at a speed of 20 ft/sec at 20 feet from 3rd. Rate of change in distance from home plate = ?

D^2 = x^2 + y^2
x = distance from 2nd to player
y = length of side (90)
D = sqrt(70^2 + 90^2)

D = 10 * sqrt(130)

2D*dD/dt = 2x*dx/dt + 2y * dy/dt

D*dD/dt = x*dx/dt

(10*sqrt(130))*dD/dt = 70*20

dD/dt = 1200 / (10*sqrt(130))

This isn't right. What went wrong?
 
Modigliani said:
Player running between 2nd base and 3rd at a speed of 20 ft/sec at 20 feet from 3rd. Rate of change in distance from home plate = ?

D^2 = x^2 + y^2
x = distance from 2nd to player
y = length of side (90)
D = sqrt(70^2 + 90^2)

D = 10 * sqrt(130)

2D*dD/dt = 2x*dx/dt + 2y * dy/dt

D*dD/dt = x*dx/dt

(10*sqrt(130))*dD/dt = 70*20

dD/dt = 1200 / (10*sqrt(130))<<< How's that - also dD/dt should be negative as the distance between the player and the HP is decreasing.

This isn't right. What went wrong?

Distance of HP to 3B = 90

Distance of player to 2B = 90 - x ---------------------------misread definition of 'x' -----------edit

Distance of player to 3B = x (dx/dt = -20)------------------------------edit

Distance of player to HP = D = sqrt[(90)^2 + (x)^2]------------------edit

Now continue....
 
Try it this way.

\(\displaystyle D^{2}=90^{2}+x^{2}\)

\(\displaystyle x=20\)

\(\displaystyle \frac{dx}{dt} = -20\)...it is negative because the distance is decreasing.

\(\displaystyle 2D\frac{dD}{dt}=2x\frac{dx}{dt}\Rightarrow \frac{dD}{dt}=\frac{x}{D}\frac{dx}{dt}\)

When x=20: \(\displaystyle D=\sqrt{90^{2}+20^{2}}=10\sqrt{85}\)

\(\displaystyle \frac{dD}{dt}=\frac{20}{10\sqrt{85}}(-20)=\frac{-8\sqrt{85}}{17}\approx -4.34\)
 

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Subhotosh Khan said:
Modigliani said:
Player running between 2nd base and 3rd at a speed of 20 ft/sec at 20 feet from 3rd. Rate of change in distance from home plate = ?

D^2 = x^2 + y^2
x = distance from 2nd to player
y = length of side (90)
D = sqrt(70^2 + 90^2)

D = 10 * sqrt(130)

2D*dD/dt = 2x*dx/dt + 2y * dy/dt

D*dD/dt = x*dx/dt

(10*sqrt(130))*dD/dt = 70*20

dD/dt = 1200 / (10*sqrt(130))<<< How's that - also dD/dt should be negative as the distance between the player and the HP is decreasing.

This isn't right. What went wrong?

Distance of HP to 3B = 90

Distance of player to 2B = x

Distance of player to 3B = 90 - x

Distance of player to HP = D = sqrt[(90)^2 + (90-x)^2]

Now continue....

What's dy/dt? -1?
 
Modigliani said:
Subhotosh Khan said:
Modigliani said:
Player running between 2nd base and 3rd at a speed of 20 ft/sec at 20 feet from 3rd. Rate of change in distance from home plate = ?

What's dy/dt? -1?

No -- is 'y' (distance between HP and 3B - length of side (90) - defined by you) changing?
 
galactus said:
Try it this way.

\(\displaystyle D^{2}=90^{2}+x^{2}\)

\(\displaystyle x=20\)

\(\displaystyle \frac{dx}{dt} = -20\)...it is negative because the distance is decreasing.

\(\displaystyle 2D\frac{dD}{dt}=2x\frac{dx}{dt}\Rightarrow \frac{dD}{dt}=\frac{x}{D}\frac{dx}{dt}\)

When x=20: \(\displaystyle D=\sqrt{90^{2}+20^{2}}=10\sqrt{85}\)

\(\displaystyle \frac{dD}{dt}=\frac{20}{10\sqrt{85}}(-20)=\frac{-8\sqrt{85}}{17}\approx -4.34\)
So basically dy/dt is zero because the side length is constant? That's much easier. Thanks.
 
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