A baseball diamondhas the shape of a square with sides 90 feet long. a player running from second base
to thrid base at a speed of 28 feet per second is 30 feet from thrid base. At what rate is the players
distance s from home plate changing?
Answer
Draw the baseball diamond, with
x=players distance from 3rd base,
y=players distance from home plate.
Then the Pythagorean theorem gives:
x^2 + 90^2 = y^2.
Differentiating both sides of the equation with respect to t gives:
2x(dx/dt) = 2y(dy/dt) (*).
When the player is 30' from 3rd base,
x=30,
y=sqrt(30^2 + 90^2)
Also, we are told that
dx/dt=-28.
Just plug in these values for x,y and dx/dt into (*) to determine dy/dt. (I'll let you do the arithmetic.)
For the baseball diamond shown in the figure below, suppose the player is running from first to second at
a speed of 27 feet per second. Find the rate at which the distance from home plate is changing when the
player is 34 feet from second base. (Round to two decimal places.) The distance between each base is
90 ft
Answer
Since the distance between each base is 90', he is 90-34=56' from 1st base.
The right triangle involved is then 90² + x²(t) = c²(t)
where x(t) is the distance he's run from 1st.
The other leg is home to 1st, so c(t) is the hypoteneuse of the triangle.
Take the derivative of that equation and note the following:
x(t) is known to be 56',
c(t) is √(90²+x²(t)) = √(8,100+3,136) = (√11,236)'.
x'(t) = 34'.
The derivative is 2x(t)x'(t) = 2c(t)c'(t), so c'(t) = x(t)x'(t)/c(t).
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the
dock that is 1 m higher than the bow of the boat.
If the rope is pulled in at a rate of 1.2 m/s, how fast is the boat approaching the dock when it is 9 m from
the dock?
2. A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first
base with a speed of 20 ft/s.
At what rate is his distance from second base changing when he is halfway to first base?
Answer = ft/s
At what rate is his distance from third base changing at the same moment?
Answer = ft/s
Answer
Baseball
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the
dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1.2 m/s, how fast is
the boat approaching the dock when it is 9 m from the dock?
Let x be the horizontal distance, y be the vertical distance,
and r bet the distance between the boat and the pulley.
We know that x = 9, y = 1, and r² = x²+y².
Also, we know that only x and r are changing, y is a constant.
In the equation, we can say that 2r dr = 2x dx
since y is a constant.
Putting in what x and r have the value of gives us
2√101 dr = 2*9 dx.
We know that dr is the rate at which the rope is being pulled.
Since the rope is being pulled in, the change is negative.
This says that dr = -1.2.
Cancelling the 2's in the equation and dividing by 9 gives us
(√101)(-1.2)/9 = dx. Evaluate this, and that gives a negative value for dx since x is decreasing. The
question, however, wants the negative of this since it asks for how fast the distance is decreasing.
2. A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first
base with a speed of 20 ft/s.
We are given a baseball diamond, which is really a square that is tipped 45°.
We know that y = 90, x = 45, dx = -20.
At what rate is his distance from second base changing when he is halfway to first base?
Answer = ft/s
Note that z² = x² + y², so 2z dz = 2x dx since y is a constant.
This means that dz = (x/z) dx.
Now z is known to be √(45²+90²) = 45√(1²+2²) = 45√5.
The value for x was given as 45.
This means that dz = (45/45√5) dx = (1/√5) dx = (√5/5)dx.
We know that dx = -20, so dz = (√5/5)(-20) = -4√5.
At what rate is his distance from third base changing at the same moment?
Answer = ft/s
Looking at the triangle, everything is symmetric between the upper and lower triangle,
so it is a negative of the other one and is therefore 4√5.