A baseball player hits a line drive to center field. As he rounds second base, he heads directly for third, running at 20 ft/sec. How fast is the distance from the rnner to home plate changing when he is halfway to third base?
aargh, did this yesterday, going to look
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h^2 = 90^2 + x^2
h is hypotenuse
x is distance from third
where x = (45-20t) so dx/dt = -20
2 h dh/dt = 2 x dx/dt
dh/dt = (x/h) (-20)
at t = 0, x = 45 and h = 100.6
so
dh/dt = -8.94
http://www.jiskha.com/display.cgi?id=1293827209
x^2 = (45-20*t)^2 + 90^2
2 x dx/dt = 2 *(45-20t) (-20)
at x=sqrt(45^2+90^2) when t = 0
2 (100.6) dx/dt = 90*-20
dx/dt = -8.94 ft/s
To determine the rate at which the distance from the runner to home plate is changing, we can use the concept of related rates.
Let's assign some variables:
- Let x represent the distance of the runner from home plate.
- Let y represent the distance of the runner from third base.
Since we are interested in the specific moment when the runner is halfway to third base, we can set up the following relationship using the Pythagorean theorem:
x^2 + y^2 = (90 ft)^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we want to find how fast the distance from the runner to home plate is changing when he is halfway to third base, we need to find dx/dt when y = 45 ft.
We have the following information:
- dx/dt = 20 ft/sec (given)
- y = 45 ft (halfway to third base)
Plugging these values into our equation, we can solve for dy/dt:
2x(20 ft/sec) + 2(45 ft)(dy/dt) = 0
Multiplying and rearranging the equation, we have:
40x + 90(dy/dt) = 0
Substituting x from the Pythagorean theorem equation, we get:
40(√[(90 ft)^2 - y^2]) + 90(dy/dt) = 0
We can now solve for dy/dt to find how fast the distance from the runner to home plate is changing when he is halfway to third base.