Rate of decreasing shadow problem solved with derivative and similar triangles

Distance of Boat Solution
Known values given: y = 8 $\frac{dz}{dt}$ = -2 $\frac{dx}{dt}$ = -2 $\frac{4}{15}$ = $\frac{-34}{15}$ Use Pythagorean Theorem Solve for z z² = x² + y² Substitute: y = 8 z² = x² - 64 Take derivative 2z $\frac{dz}{dt}$ = 2x $\frac{dx}{dt}$ Substitute: $\frac{dx}{dt}$ = $\frac{-34}{15}$ & $\frac{dz}{dt}$ = -2 2z * -2 = 2x * $\frac{-34}{15}$ -4z = $\frac{-68x}{15}$ z = $\frac{-68x}{-60}$ = $\frac{17x}{15}$ Use Pythagorean Theorem Solve for x x² = z² - y² Substitute: z = $\frac{17x}{15}$ & y = 8 x² = ( $\frac{17x}{15}$ )² – 8² $\frac{225x²}{225}$ = $\frac{289x²}{225}$ - 64 $\frac{225x²}{225}$ - $\frac{289x²}{225}$ = -64 $\frac{-64x²}{225}$ = -64 x² = $\frac{-225}{64}$ * -64 x² = 225 $\sqrt{x²}$ = $\sqrt{225}$ x = 15 The boat is 15 feet from the shore
156C

Distance of Boat
Solution

Known values given:
y = 8 $\frac{dz}{dt}$ = -2 $\frac{dx}{dt}$ = -2 $\frac{4}{15}$ = $\frac{-34}{15}$
Drawing of water filling a cone with r and h coordinates

Use Pythagorean Theorem
Solve for z
z² = x² + y²

Substitute: y = 8
z² = x² - 64

Take derivative
2z $\frac{dz}{dt}$ = 2x $\frac{dx}{dt}$

Substitute: $\frac{dx}{dt}$ = $\frac{-34}{15}$ & $\frac{dz}{dt}$ = -2
2z * -2 = 2x * $\frac{-34}{15}$
-4z = $\frac{-68x}{15}$
z = $\frac{-68x}{-60}$ = $\frac{17x}{15}$

Use Pythagorean Theorem
Solve for x
x² = z² - y²

Substitute: z = $\frac{17x}{15}$ & y = 8
x² = ( $\frac{17x}{15}$ )² – 8²
$\frac{225x²}{225}$ = $\frac{289x²}{225}$ - 64
$\frac{225x²}{225}$ - $\frac{289x²}{225}$ = -64
$\frac{-64x²}{225}$ = -64
x² = $\frac{-225}{64}$ * -64
x² = 225
$\sqrt{x²}$ = $\sqrt{225}$
x = 15

The boat is 15 feet from the shore

156C

Distance of BoatSolution

Distance of Boat
Solution