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[SOLVED] Simple Related Rates
A baseball diamond is a square 90 ft on a side. A runner travels from home plate to first base at 20 ft/sec. How fast is the runner's distance changing when the runner is half way to first base?
a^2+b^2=c^2
The distance from the runner to second base is: D=sqrt((20t)^2+90^2)
dD/dt=(40t)/(sqrt(4x^2+81))
The runner is half way to second base when... 20t=45; t=9/4;
I plug t into dD/dt and get 4*sqrt(5)~=8.94. But wait...shouldn't dD/dt be negative, because D is decreasing, because the runner is getting closer to second base?
EDIT: I got it. Distance between runner and first base=r=90-20t. Distance between runner and second base=D=sqrt(r^2+90^2)
dD/dt=(1/2)(r^2+90^2)^(-1/2) * (2r*dr/dt)
Like I said, the runner is half way to second base when 20t=45;t=9/4.
dr/dt=-20
r=90-20(9/4)
Plug r and dr/dt in, and get -4*sqrt(5), which I'm pretty sure is the correct answer.
Homework Statement
A baseball diamond is a square 90 ft on a side. A runner travels from home plate to first base at 20 ft/sec. How fast is the runner's distance changing when the runner is half way to first base?
Homework Equations
a^2+b^2=c^2
The Attempt at a Solution
The distance from the runner to second base is: D=sqrt((20t)^2+90^2)
dD/dt=(40t)/(sqrt(4x^2+81))
The runner is half way to second base when... 20t=45; t=9/4;
I plug t into dD/dt and get 4*sqrt(5)~=8.94. But wait...shouldn't dD/dt be negative, because D is decreasing, because the runner is getting closer to second base?
EDIT: I got it. Distance between runner and first base=r=90-20t. Distance between runner and second base=D=sqrt(r^2+90^2)
dD/dt=(1/2)(r^2+90^2)^(-1/2) * (2r*dr/dt)
Like I said, the runner is half way to second base when 20t=45;t=9/4.
dr/dt=-20
r=90-20(9/4)
Plug r and dr/dt in, and get -4*sqrt(5), which I'm pretty sure is the correct answer.
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