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 July 23rd, 2012, 07:03 PM #1 Senior Member   Joined: Jun 2011 Posts: 154 Thanks: 0 Rate of change of distance If anyone could help me with this problem, I'd greatly appreciate it. I honestly don't even know how to start this one, can't find any similar examples in my notes, and will need to do one just like this on a test tomorrow. Any help is greatly appreciated. A (square) baseball diamond has sides that are 90 feet long. A player 20 feet from third base is running at a speed of 21 feet per second. At what rate is the player's distance from home plate changing? (Round your answer to two decimal places.)
July 23rd, 2012, 07:58 PM   #2
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From: St. Augustine, FL., U.S.A.'s oldest city

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Math Focus: Calculus/ODEs
Re: Rate of change of distance

I drew a rough sketch:

[attachment=0:2oi6r39t]baseballplayer.jpg[/attachment:2oi6r39t]

The player is at point P and his velocity has been resolved into its x and y components to the right.

The player is running along the line $y=x+45\sqrt{2}$ and his distance D from home plate H is:

$D=\sqrt{(x-0)^2+$$y-\(-45\sqrt{2}$$\)^2}$

Substitute for y:

$D=\sqrt{x^2+$$x+45\sqrt{2}+45\sqrt{2}$$^2}=\sqrt{2 x^2+180\sqrt{2}x+16200}$

Differentiate with respect to time t:

$\frac{dD}{dt}=\frac{2x+90\sqrt{2}}{\sqrt{2x^2+180\ sqrt{2}x+16200}}\cdot\frac{dx}{dt}$

When the player is 20 feet from 3rd base, his x-coordinate is $-\frac{7}{9}\cdot45\sqrt{2}=-35\sqrt{2}$ and we have:

$\frac{dx}{dt}=-\frac{21}{\sqrt{2}}\:\text{\frac{ft}{s}}$ hence:

$\frac{dD}{dt}\|_{\small{x=-35\sqrt{2}}}=\frac{2$$-35\sqrt{2}$$+90\sqrt{2}}{\sqrt{2$$-35\sqrt{2}$$^2+180\sqrt{2}$$-35\sqrt{2}$$+16200}}$$-\frac{21}{\sqrt{2}}\:\text{\frac{ft}{s}}$$=-\frac{420}{10\sqrt{85}}\:\text{\frac{ft}{s}}=-\frac{42}{\sqrt{85}}\:\text{\frac{ft}{s}}$
Attached Images
 baseballplayer.jpg (16.6 KB, 264 views)

 July 24th, 2012, 03:09 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rate of change of distance A simpler approach: Orient the diamond such that 3rd base is at the origin and home plate is at (0,-90) and the player is then running along the x-axis. Suppose the player is at (x,0) and we are told $\frac{dx}{dt}=-21\text{ \frac{ft}{s}}$ The distance between the player and home plate is therefore: $D=\sqrt{x^2+90^2}$ hence, differentiating with respect to t: $\frac{dD}{dt}=\frac{x}{\sqrt{x^2+90^2}}\cdot\frac{ dx}{dt}$ When the player is 20 ft from home base, i.e., x = 20, we find: $\frac{dD}{dt}\|_{\small{x=20}}=\frac{20}{\sqrt{20^ 2+90^2}}$$-21\text{ \frac{ft}{s}}$$=-\frac{42}{\sqrt{85}}\:\text{\frac{ft}{s}}$

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