6. a baseball diamond is a square 90 ft. on each side. a player is running to the first base at the rate of…

Explanation:

Step1: Establish a right - triangle relationship

Let the side of the baseball diamond be (a = 90) ft. Let (x) be the distance of the player from home - plate and (y) be the distance of the player from second base. By the Pythagorean theorem, (y^{2}=(90)^{2}+(90 - x)^{2}).

Step2: Differentiate both sides with respect to time (t)

Using the chain - rule, (2y\frac{dy}{dt}=2(90 - x)(- \frac{dx}{dt})). Then (\frac{dy}{dt}=\frac{(x - 90)\frac{dx}{dt}}{y}).

Step3: Find the values of (x), (y) and (\frac{dx}{dt})

The player runs at a rate (\frac{dx}{dt}=30) ft/sec. When the player has run halfway, (x = 45) ft. Then (y=\sqrt{90^{2}+(90 - 45)^{2}}=\sqrt{8100 + 2025}=\sqrt{10125}=45\sqrt{5}) ft.

Step4: Substitute the values into the derivative formula

Substitute (x = 45), (y = 45\sqrt{5}) and (\frac{dx}{dt}=30) into (\frac{dy}{dt}=\frac{(x - 90)\frac{dx}{dt}}{y}). We get (\frac{dy}{dt}=\frac{(45 - 90)\times30}{45\sqrt{5}}=\frac{- 45\times30}{45\sqrt{5}}=- 6\sqrt{5}) ft/sec.

Answer:

b. (-6\sqrt{5}) ft/sec