Moving Ladder problem solved with derivative and Pythagorean Theorem

Moving Ladder Solution
By the Pythagorean Theorem the top of the ladder will be 12 feet from the floor. $\sqrt{13² - 5²}$ = 12. Known values: x = 5; y = 12; $\frac{dx}{dt}$ = 3 this is the rate bottom of ladder is moving. Need to find $\frac{dy}{dt}$ . This is the rate top of the ladder is moving down. Use Pythagorean Theorem formula. x²+ y² = 13² Take derivative. Solve for $\frac{dy}{dt}$ . 2x $\frac{dx}{dt}$ + 2y $\frac{dy}{dt}$ = 0 Substitute x = 5 $\frac{dx}{dt}$ = 3 y = 12 253 + 212 $\frac{dy}{dt}$ = 0 30 + 24 $\frac{dy}{dt}$ = 0 $\frac{dy}{dt}$ = $\frac{-30}{24}$ = $\frac{-5}{4}$ The top of ladder is moving down at $\frac{5}{4}$ feet per second when the bottom of the ladder is 5 feet from the wall.
152

Moving Ladder
Solution

By the Pythagorean Theorem the top of
the ladder will be 12 feet from the floor.

\sqrt{13² - 5²}

= 12.

Known values: x = 5; y = 12;

\frac{dx}{dt}

= 3 this
is the rate bottom of ladder is moving.
Need to find

\frac{dy}{dt}

. This is the rate top of
the ladder is moving down.
Drawing of ladder solution with x and y coordinated

Use Pythagorean Theorem formula.
x²+ y² = 13²

Take derivative. Solve for $\frac{dy}{dt}$ .
2x $\frac{dx}{dt}$ + 2y $\frac{dy}{dt}$ = 0

Substitute x = 5 $\frac{dx}{dt}$ = 3 y = 12
2*5*3 + 2*12* $\frac{dy}{dt}$ = 0
30 + 24 $\frac{dy}{dt}$ = 0
$\frac{dy}{dt}$ = $\frac{-30}{24}$ = $\frac{-5}{4}$

The top of ladder is moving down at

\frac{5}{4}

feet per second when the bottom of
the ladder is 5 feet from the wall.

152

Moving LadderSolution

Moving Ladder
Solution