# A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s. a) At what rate is his distance from second base decreasing when he is...

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 24 ft/s.

a) At what rate is his distance from second base decreasing when he is halfway to first base.

b) At what rate is his distance from third base increasing at the same moment?

### Textbook Question

Chapter 3, 3.9 - Problem 18 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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electreto05 | College Teacher | (Level 1) Assistant Educator

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In both cases we must find the rate of change of the hypotenuse of a right triangle. In part (a), the triangle is formed between the first base, second base and the runner; in part (b) the vertices are third base, home plate and the batter.

Let's name the following parameters:

s = distance between home plate and first base.

We call b = 90 ft,  the other three sides of the square.

ds/dt = v = 24 ft/s = speed when the batter is moved from the home

h = distance from second base (part a) or third base (part b) to the batter at any time. It is the hypotenuse of our triangles.

According to the theorem of Pythagoras, the hypotenuse h is:

h2 = s2 + b2

h = (s2 + b2)1/2

a)

To find the rate of decrease of h we derive this expression with respect to time:

-(dh/dt) = 1/2(s2 + b2)-1/22s(ds/dt); the minus sign expresses the decrease of h when t increases.

dh/dt = -[1/2(s2 + b2)-1/22s(v)] = -[ sv/(s2 + b2)1/2]

Now we substitute the given values:

dh/dt = -[(s/2)v/((s/2)2 + b2)1/2]= -[(45)(24)/√(2025 + 8100)]

dh/dt = -10.73 ft/s

b)

The procedure for finding the increase of h from third base is the same, but considering the positive sign for the derivative dh/dt. So in this case we have:

dh/dt = 10.73 ft/s

Here is an example with some different numbers for a similar problem.

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

(a) The distance from second base can be found using the Pythagorean Theorem.

`d^2=90^2+(90-24t)^2`

(t is the time in seconds from the hit -- 90-24t is the distance the runner is from first base at time t seconds.)

So:

`d^2=16200-4320t+576t^2`

Differentiate with respect to t to get:

`2d (dd)/(dt)=-4320+1152t`

When the runner is half way to first base, 2 seconds have elapsed. Also, the distance from first is 48 feet, so d=102 and t=2. Substituting we get:

`(dd)/(dt)=(-4320+1152(2))/(2*102)`

So the rate of change at this point is approximately -9.88 feet per second.

(b) The distance from third base can also be found using the Pythagorean Theorem:

`d^2=90^2+(24t)^2`

`d^2=8100+576t^2`

Differentiating with respect to t we get:

`2d (dd)/(dt)=1152t`

`(dd)/(dt)=(1152t)/(2d)`

Again, t=2 and d=102 so:

`(dd)/(dt)=(1152(2))/(2(102))~~11.29` feet per second.

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