QuestionLadder on wall: top slips down as base moves away at 5 m/s. Find (29) rate of top when 5 m from ground, and (30) rate of area change when top is 5 m from ground.

Studdy Solution

STEP 1

1. The ladder, wall, and ground form a right triangle with the ladder as the hypotenuse.
2. The rate at which the base of the ladder moves away from the wall is constant at $5 \mathrm{~m/s}$ .
3. The Pythagorean theorem can be used to relate the lengths of the sides of the triangle.
4. Related rates can be used to find the rate at which the top of the ladder moves down the wall and the rate at which the area of the triangle changes.

STEP 2

1. Use the Pythagorean theorem to relate the lengths of the sides of the triangle.
2. Differentiate the Pythagorean theorem with respect to time to find the rate at which the top of the ladder is moving down the wall.
3. Use the formula for the area of a right triangle to relate the area to the sides of the triangle.
4. Differentiate the area with respect to time to find the rate at which the area is changing when the top of the ladder is 5 meters from the ground.

STEP 3

Write the Pythagorean theorem for the right triangle formed by the ladder, wall, and ground.
$x^2 + y^2 = l^2$
where $x$ is the distance from the base of the ladder to the wall, $y$ is the height of the ladder on the wall, and $l$ is the length of the ladder.

STEP 4

Differentiate the Pythagorean theorem with respect to time $t$ .
$\frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(l^2)$
Since the length of the ladder $l$ is constant, $\frac{d}{dt}(l^2) = 0$ .

STEP 5

Apply the chain rule to differentiate $x^2$ and $y^2$ .
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

STEP 6

Solve for $\frac{dy}{dt}$ when $y = 5$ meters and $\frac{dx}{dt} = 5 \mathrm{~m/s}$ .
$2x(5) + 2(5)\frac{dy}{dt} = 0$

STEP 7

Use the Pythagorean theorem to find $x$ when $y = 5$ meters and $l = 13$ meters.
$x^2 + 5^2 = 13^2$

STEP 8

Solve for $x$ .
$x^2 = 13^2 - 5^2$

STEP 9

Calculate $x$ .
$x = \sqrt{13^2 - 5^2}$

STEP 10

Substitute $x$ back into the differentiated Pythagorean theorem to solve for $\frac{dy}{dt}$ .
$2(\sqrt{13^2 - 5^2})(5) + 2(5)\frac{dy}{dt} = 0$

STEP 11

Calculate $\frac{dy}{dt}$ .
$\frac{dy}{dt} = -\frac{2(\sqrt{13^2 - 5^2})(5)}{2(5)}$

STEP 12

Simplify to find the rate at which the top of the ladder is moving down the wall.
$\frac{dy}{dt} = -\sqrt{13^2 - 5^2}$

STEP 13

Write the formula for the area $A$ of the right triangle formed by the ladder, wall, and ground.
$A = \frac{1}{2}xy$

STEP 14

Differentiate the area with respect to time $t$ .
$\frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right)$

STEP 15

Substitute the known values for $x$ , $y$ , $\frac{dx}{dt}$ , and $\frac{dy}{dt}$ into the differentiated area formula.
$\frac{dA}{dt} = \frac{1}{2}\left(\sqrt{13^2 - 5^2}(-\sqrt{13^2 - 5^2}) + 5(5)\right)$

STEP 16

Calculate $\frac{dA}{dt}$ .
$\frac{dA}{dt} = \frac{1}{2}\left(-(13^2 - 5^2) + 25\right)$

STEP 17

Simplify to find the rate at which the area is changing.
$\frac{dA}{dt} = \frac{1}{2}\left(-169 + 25 + 25\right)$

STEP 18

Finish the calculation for $\frac{dA}{dt}$ .
$\frac{dA}{dt} = \frac{1}{2}\left(-144 + 25\right)$

STEP 19

Simplify the final expression for $\frac{dA}{dt}$ .
$\frac{dA}{dt} = \frac{1}{2}(-119)$

STEP 20

Provide the solutions for the rates at which the top of the ladder is moving and the area is changing when the top of the ladder is 5 meters from the ground.
The top of the ladder is moving down at a rate of $-\sqrt{13^2 - 5^2} \mathrm{~m/s}$ , and the area of the triangle is changing at a rate of $-\frac{119}{2} \mathrm{~m^2/s}$ .

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