Baseball Diamond Distance Problem

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Homework Help Overview

The problem involves a baseball diamond, which is a square with sides of 90 feet. A player is running from first base to second base at a speed of 16 ft/sec, and the question seeks to determine the rate at which the player's distance from third base is changing when the player is 30 feet from first base.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to differentiate an equation related to the distances involved, questioning the distinction between velocity and acceleration. There is an exploration of how to set up the derivative and the relationship between the variables in the context of the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the differentiation. Some participants express confusion about the notation and the setup of the equation, while others attempt to clarify the relationships between the variables involved.

Contextual Notes

There is uncertainty regarding the correct interpretation of terms such as "rate of change" and whether it refers to velocity or acceleration. Participants are also navigating the implications of differentiating the equation and the appropriate timing for substituting values.

zelphics
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Homework Statement


A baseball diamond is a square 90 feet on a side. A player runs from first base to second at a rate of 16 ft/sec. At what rate is the player's distance from third base changing when the player is 30 ft from first base..



Homework Equations


dr/dt ds/dt

dx/dt? a^2+b^2=C^2


The Attempt at a Solution



I have no idea... i don't know where to start, but it must be solved using the derivative.
 
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hint: they are asking for acceleration when they say "at what rate... blah blah blah... changing"

and there are wel known aceleation, distance, time equations out there
 
wouldn't they be asking for velocity or rate, not acceleration?
 
when there is a rate of change that is synonymous with acceleration
 
They're asking for the players rate of change of distance, which infers velocity, as opposed to acceleration, which is the rate of change of velocity.

You have your equation set up correctly, why don't you begin by differentiating both sides.
 
oops. my bad.

change of rate wrt distance = velocity
change of rate wrt time = acceleration

so sue me.
 
i don't care that you were wrong, i just have no idea where to start? and I am not sure i understand the notation..
 
Do you know how to differientiate both sides of the equation you posted? You'll want to include the derivative notation for each variable when you do this.

For example, if I had 2x^2 = c^3 and I differientiated both sides, I would have 4x \frac{dx}{dt} = 3c^2 \frac{dc}{dt}.
 
i know how to differentiate both sides, but i don't even have two sides...
 
  • #10
In your opening post, you had the equation "a^2+b^2=C^2," right? There are two sides to this equation.
 
  • #11
i don't understand how that works? but it would be:

60^2 + 90^2 = C^2
 
  • #12
You'll want to wait until the very end to plug in your values. Just begin by treating "a," "b," and "c" as variables, and take the derivative of both sides of the equation with respect to time.
 
  • #13
2a + 2b = 2c?
 
  • #14
it seems like the type of problem where first you do a derivative

and then once you have a resultant equation you plug in 30 for one of the variables
 
  • #15
zelphics said:
2a + 2b = 2c?

This is correct, and sprint is right, you'll want to wait until the very end to plug in values. Your equation is missing three things however, you'll want to include the da/dt, db/dt, and dc/dt terms next to their respective variables. So far so good?
 
Last edited:
  • #16
yea so da/dt(2a) + db/dt(2b) = dc/dt(2c)?
 
  • #17
Right, so now the 2's cancel out. The problem said that you're dealing with two sides of a square, so how can we relate "a" and "b" ? What's our new equation?
 

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