Related Rates: Baseball Diamond and Fish Reeling

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SUMMARY

The discussion focuses on solving related rates problems involving a baseball diamond and a fish being reeled in. The first problem involves a player running from first base to second base, with the correct application of the Pythagorean Theorem yielding a rate of change of -37.5 ft/s for the distance from home plate. The second problem, concerning a fish being reeled in at 2 m/s, results in a rate of -2.5 m/s for the fish's movement through water. The negative values indicate a decrease in distance, which is correctly interpreted in the context of the problems.

PREREQUISITES
  • Understanding of related rates in calculus
  • Proficiency in applying the Pythagorean Theorem
  • Familiarity with differentiation and rates of change
  • Knowledge of interpreting negative values in the context of motion
NEXT STEPS
  • Study the application of the Pythagorean Theorem in related rates problems
  • Learn about differentiating implicit functions in calculus
  • Explore more complex related rates scenarios involving multiple variables
  • Review the concept of velocity and speed, focusing on magnitude versus direction
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Students studying calculus, particularly those focusing on related rates, as well as educators looking for examples to illustrate these concepts in a practical context.

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Can someone check if i did these two questions right please?

The questions are:

1. A baseball diamond is a perfect square with each side measuring 90 feet
in length. A player runs from the first base to second base at a speed of
25 ft/s. How fast is she moving from home plate when she is 30 ft away from
first base.

2. A fish is being reeled in a rate of 2 m/s (i.e, the fishing line is
being shortened by 2 m/s by a man. If he is sitting 30 m above water on a
dock, how fast is the fish moving through water when the line is 50 m long.

My answers are these:

1. A drew my diamond and figured that i will be using half of the diamond.

Let the distance between home plate and first base be x
Let the distance from first base to second base by y

Given
y = 90 ft
dy/dt = 25 ft/s
x = 90 ft - 30ft
=60 ft
dx/dt = ?

x^2 + y^2 = z^2
90^2 + 90^2 = z^2
z = 127.3 ft

x^2 + y^2 = z^2
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
sine dz/dt = 0
2(60)(dx/dt) + 2(90)(25) = 0
dx/dt = -37.5 ft/s

Is this correct? and why is my answer negative shouldn't it be positive
because you increasing the distance from home plate or is my answer
negative because the distance to first base is decreasing.


2. Given
y = 30 m
z = 50 m
dz/dt = - 2m/s
x = ?
dx/dt = ?

x^2 + y^2 = z^2
x^2 = 50^2 - 30^2
x = 40 m

x^2 + y^2 = z^2
2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)
since dy/dt = 0
2(40)(dx/dt) = 2(50)(-2)
dx/dt = -2.5 m/s

Is this question correct because I am not sure when to make my numbers
negative, i know we have to when something is decreasing but I am not sure.

So can you please check if i did this two questions right?
 
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For the first one, y isn't 90. Also, you won't find x by subtracting. It's a right triangle. How can you find an unknown side of a right triangle? The baseball player is 30 ft. from second base, not 90 ft. You had the general setup right though.

Using your defined variables.

x^2=90^2+y^2

x\frac{dx}{dt}=y\frac{dy}{dt}

You know y, dy/dt, and you can solve for x using the Pythagorean Theorem. Now just find dx/dt.
 
The second one looks good to me. Your sign is correct. This question is worded weirdly. It asks how fast the fish is moving through the water and I interpret that to mean what speed, which is |\vec{v}|, where v is velocity. I think going by the question it's more correct to put the positive value, only showing magnitude and not direction, but you should ask your teacher.
 

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