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Question 3

Page history last edited by Rence 15 years ago

A Calculator MAY be used to help answer this question.

Be certain to limit yourself to the four permissible calculator functions.



A baseball diamond is a square with each side 90 feet in length. A player runs from second base to third base at a rate of 18 ft/sec.


(a) At what rate is the player's distance from first base, A, changing when his distance from third base, D, is 22.5 feet?


(b) At what rate is angle α increasing when D is 22.5 feet?


(c) At what rate is the area of the trapezoidal region, formed by segments A, B, C and D changing when D is 22.5 feet?


Quick Find Code For This Problem.


Legend: [QUES3LGND]

Solution For Part a.): [QUES3PA]

Solution For Part b.): [QUES3PB]

Solution For Part c.): [QUES3PC]

Final Answers All Grouped Together: [QUES3ANS]


How To Use Quick Find.


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  4. Numerically integrating a function: [2nd] [7:∫f(x)dx] or [MATH] [9:fnInt]


Welcome to question 3. I am Francis and I will be solving this question.

Start by drawing a diagram of the white triangle found in the baseball diamond.




Each side of the baseball diamond is 90 ft. We use this information to find side S, due to the player being 22.5 ft from third base. We know he is 90 - 22.5 = 67.5 ft from second base, so S = 67.5 ft. Now that we know 2 of the 3 sides of this triangle, we can use the pythagorean theorm to find side A.




A  = 112.5 ft.        Do not forget the units.


We are given the initial rate at which the player runs from 2nd to 3rd base(dS/dt = 18 ft/sec.). The rate will be positive because the distance bewteen the player and 2nd base is increasing. We have to find the rate side A is changing while the runner is 22.5 ft. from third base, or in algebraic terms: find dA/dt when A = 112.5 ft.


We know A² - S2 = 8100 (modified Pythagorean Theorem) when A = 112.5 ft. We differentiate this equation implicitly to get the rate, dA/dt.



We know dS/dt = 18 because it was given. We know side A = 112.5 ft, and side S = 67.5 ft.



dA/dt = 10.8


Therefore side A is increasing at a rate of 10.8 ft/sec, when the player is running at 18 ft/sec, and is 22.5 ft, from third base.


PART A is done.





Using the triangle from above, we can see the position of the angle. We have to find the rate of this angle when D is at 22.5 ft.



We know the length of all sides of the triangle, so we can use either tangent or cosine to find angle α. I prefer tangent, because it makes use of the constant, which is always safer than compounding errors.



tanα = 0.75

Angle α = 0.6435 radians


Now we find dα/dt when α = 0.6435


We relate angle α to side S to find dα/dt.



Where 90 is a constant. We differentiate this and we get:



We know dS/dt is 18 ft/sec. and    and α = 0.6435



dα/dt = 0.128 radians/sec.

Angle α is increasing at a rate of 0.128 radians/sec. when side D is at 22.5 ft.


PART B is done.



 Part C.

We have to find rate at which the area of the trapezoid is changing when D is 22.5 ft. Here's a diagram and equation for the area of a trapezoid:



Where A is the area, h is the perpendicular distance from the two parallel sides, and b1 and b2 are the lengths of the 2 parallel sides. This formula is derived from the sum of the area of triangles and the area of a rectangle/square. Here is a diagram:



So inputing the labels from the diagram in the question we get:





Where T is the area of the trapezoid, C is the length from 3rd base to Home and B is the length from Home to 1st Base.

Now we try to relate an equation to the area. We know C and B are constants that each equal 90 ft.

We can use this to find an equation to relate the area T to one of the sides and differentiate it.

Using the constants and variables we find this equation.



Factor in the 90.



Divide both terms in the numerator by 2.


T(D) = 45D + 4050


We now have an equation that relates T as a function of D, which can be used to find dT/dt.


To find dT/dt, we differentiate T(D) and we get:



We have not yet found the value of the rate dD/dt, but we know that dS/dt = 18 ft/sec. and because they are related to the same side (bases 2nd to 3rd) and the player is running 18 ft/sec away from 2nd base, we know dD/dt = -18ft/sec.

Now we can find dT/dt since all values are known:



dT/dt = -810 ft2/sec.

This value is negative because the area of the trapezoid is decreasing when side D is decreasing because the player runs to 3rd base.


The rate at which the area of the trapezoidal region is changing is -810 ft2/sec. when side D = 22.5 ft.


Part C is done.




A) dA/dt = 10.8 ft/sec.

B) dα/dt = 0.128 radians/sec.

C) dT/dt = -810 ft2/sec.


This is how you solve question 3.


Comments (2)

Francis said

at 7:05 pm on Apr 17, 2009

How do I add pictures from my computer?

Benchmen said

at 7:40 pm on Apr 17, 2009

While you are editing you will see a tab to the right that says add images or something. Once you click on there you can upload by clicking upload.

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