Answer
(a) His distance from second base is decreasing at a rate of $10.7~ft/s$
(b) His distance from third base is increasing at a rate of $10.7~ft/s$
Work Step by Step
(a) Let $x$ be the distance between the batter and first base:
Then $x = 45~ft$
Let $y$ be the distance between second base and first base:
Then $y = 90~ft$
Let $z$ be the distance between the batter and second base. We can find $z$ when the batter is halfway to first base:
$z^2 = x^2+y^2$
$z = \sqrt{x^2+y^2}$
$z = \sqrt{(45~ft)^2+(90~ft)^2}$
$z = 100.623~ft$
We can differentiate both sides of the equation with respect to $t$:
$z^2 = x^2+y^2$
$2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$
$\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$
$\frac{dz}{dt} = \frac{1}{100.623}~[(45)(-24) + (90)(0)]$
$\frac{dz}{dt} = -10.7~ft/s$
His distance from second base is decreasing at a rate of $10.7~ft/s$
(b) Let $x$ be the distance between the batter and home plate:
Then $x = 45~ft$
Let $y$ be the distance between third base and home plate:
Then $y = 90~ft$
Let $z$ be the distance between the batter and third base. We can find $z$ when the batter is halfway to first base:
$z^2 = x^2+y^2$
$z = \sqrt{x^2+y^2}$
$z = \sqrt{(45~ft)^2+(90~ft)^2}$
$z = 100.623~ft$
We can differentiate both sides of the equation with respect to $t$:
$z^2 = x^2+y^2$
$2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$
$\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$
$\frac{dz}{dt} = \frac{1}{100.623}~[(45)(24) + (90)(0)]$
$\frac{dz}{dt} = 10.7~ft/s$
His distance from third base is increasing at a rate of $10.7~ft/s$
