Math / Calculus

QuestionRemaining Time: 09:03:27
A ladder 17 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of $8 \mathrm{ft} / \mathrm{s}$ , how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 2 ft from the wall?
Do not include units in your answer. Your answer can be exact or approximate. If it is approximate, round to three decimal places.

Studdy Solution

STEP 1

What is this asking? How fast is the top of a 17-foot ladder moving down a wall if its bottom is sliding away at 8 ft/s when the bottom is 2 feet away from the wall? Watch out! The ladder's length is constant, but the triangle it forms with the wall and ground is changing!
Also, we're dealing with rates of change, so think derivatives!

STEP 2

1. Set up the relationship between the ladder, wall, and ground.
2. Differentiate with respect to time.
3. Plug in the known values and solve for the rate of change of the top of the ladder.

STEP 3

Imagine a right triangle: the ladder is the hypotenuse ( $L = 17$ ), the wall is the vertical side ( $y$ ), and the ground is the horizontal side ( $x$ ).

STEP 4

We know $x^2 + y^2 = L^2$ .
Since $L = 17$ , we have $x^2 + y^2 = 17^2 = 289$ .
This equation connects the positions of the top and bottom of the ladder.

STEP 5

Differentiating both sides with respect to time ( $t$ ) gives us $2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$ .
Remember, $x$ and $y$ are functions of time, so we use the chain rule!

STEP 6

We can simplify by dividing both sides of the equation by 2, giving us $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0$ .
This makes the numbers easier to work with!

STEP 7

We're given $\frac{dx}{dt} = 8$ (the rate at which the bottom slides away) and $x = 2$ .
We need to find $\frac{dy}{dt}$ (how fast the top is sliding down).

STEP 8

When $x = 2$ , we can use $x^2 + y^2 = 289$ to find $y$ .
So, $2^2 + y^2 = 289$ , meaning $4 + y^2 = 289$ , so $y^2 = 285$ .
Since $y$ represents a length, we take the positive square root: $y = \sqrt{285}$ .

STEP 9

Plugging in the values into $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0$ , we get $2 \cdot 8 + \sqrt{285} \cdot \frac{dy}{dt} = 0$ , which simplifies to $16 + \sqrt{285} \cdot \frac{dy}{dt} = 0$ .
Now, we isolate $\frac{dy}{dt}$ : $\frac{dy}{dt} = -\frac{16}{\sqrt{285}}$ .

STEP 10

Calculating the fraction gives us $\frac{dy}{dt} \approx -0.953$ .
Since the question asks us not to include units, and to round to three decimal places if approximating, this is our final answer.

STEP 11

The top of the ladder is sliding down the wall at approximately $-0.953$ ft/s.
The negative sign indicates that the top is moving downwards.

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Studdy Solution

STEP 1

STEP 2

1. Set up the relationship between the ladder, wall, and ground.
2. Differentiate with respect to time.
3. Plug in the known values and solve for the rate of change of the top of the ladder.

STEP 3

Imagine a right triangle: the ladder is the hypotenuse ( $L = 17$ ), the wall is the vertical side ( $y$ ), and the ground is the horizontal side ( $x$ ).

STEP 4

We know $x^2 + y^2 = L^2$ .
Since $L = 17$ , we have $x^2 + y^2 = 17^2 = 289$ .
This equation connects the positions of the top and bottom of the ladder.

STEP 5

Differentiating both sides with respect to time ( $t$ ) gives us $2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$ .
Remember, $x$ and $y$ are functions of time, so we use the chain rule!

STEP 6

We can simplify by dividing both sides of the equation by 2, giving us $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0$ .
This makes the numbers easier to work with!

STEP 7

We're given $\frac{dx}{dt} = 8$ (the rate at which the bottom slides away) and $x = 2$ .
We need to find $\frac{dy}{dt}$ (how fast the top is sliding down).

STEP 8

When $x = 2$ , we can use $x^2 + y^2 = 289$ to find $y$ .
So, $2^2 + y^2 = 289$ , meaning $4 + y^2 = 289$ , so $y^2 = 285$ .
Since $y$ represents a length, we take the positive square root: $y = \sqrt{285}$ .

STEP 9

Plugging in the values into $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0$ , we get $2 \cdot 8 + \sqrt{285} \cdot \frac{dy}{dt} = 0$ , which simplifies to $16 + \sqrt{285} \cdot \frac{dy}{dt} = 0$ .
Now, we isolate $\frac{dy}{dt}$ : $\frac{dy}{dt} = -\frac{16}{\sqrt{285}}$ .

STEP 10

Calculating the fraction gives us $\frac{dy}{dt} \approx -0.953$ .
Since the question asks us not to include units, and to round to three decimal places if approximating, this is our final answer.

STEP 11

The top of the ladder is sliding down the wall at approximately $-0.953$ ft/s.
The negative sign indicates that the top is moving downwards.

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Studdy Solution

STEP 1

STEP 2

1. Set up the relationship between the ladder, wall, and ground.2. Differentiate with respect to time.3. Plug in the known values and solve for the rate of change of the top of the ladder.

STEP 3

Imagine a right triangle: the ladder is the **hypotenuse** (L=17L = 17L=17), the wall is the **vertical side** (yyy), and the ground is the **horizontal side** (xxx).

STEP 4

We know x2+y2=L2x^2 + y^2 = L^2x2+y2=L2.Since L=17L = 17L=17, we have x2+y2=172=289x^2 + y^2 = 17^2 = 289x2+y2=172=289.This equation connects the positions of the top and bottom of the ladder.

STEP 5

Differentiating both sides with respect to time (ttt) gives us 2x⋅dxdt+2y⋅dydt=02x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 02x⋅dtdx​+2y⋅dtdy​=0.Remember, xxx and yyy are functions of time, so we use the chain rule!

STEP 6

We can simplify by dividing both sides of the equation by 2, giving us x⋅dxdt+y⋅dydt=0x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0x⋅dtdx​+y⋅dtdy​=0.This makes the numbers easier to work with!

STEP 7

We're given dxdt=8\frac{dx}{dt} = 8dtdx​=8 (the rate at which the bottom slides away) and x=2x = 2x=2.We need to find dydt\frac{dy}{dt}dtdy​ (how fast the top is sliding down).

STEP 8

When x=2x = 2x=2, we can use x2+y2=289x^2 + y^2 = 289x2+y2=289 to find yyy.So, 22+y2=2892^2 + y^2 = 28922+y2=289, meaning 4+y2=2894 + y^2 = 2894+y2=289, so y2=285y^2 = 285y2=285.Since yyy represents a length, we take the positive square root: y=285y = \sqrt{285}y=285​.

STEP 9

STEP 10

Calculating the fraction gives us dydt≈−0.953\frac{dy}{dt} \approx -0.953dtdy​≈−0.953.Since the question asks us not to include units, and to round to three decimal places if approximating, this is our final answer.

STEP 11

The top of the ladder is sliding down the wall at approximately −0.953-0.953−0.953 ft/s.The negative sign indicates that the top is moving downwards.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Set up the relationship between the ladder, wall, and ground.
2. Differentiate with respect to time.
3. Plug in the known values and solve for the rate of change of the top of the ladder.

Imagine a right triangle: the ladder is the hypotenuse ( $L = 17$ ), the wall is the vertical side ( $y$ ), and the ground is the horizontal side ( $x$ ).

We know $x^2 + y^2 = L^2$ .
Since $L = 17$ , we have $x^2 + y^2 = 17^2 = 289$ .
This equation connects the positions of the top and bottom of the ladder.

Differentiating both sides with respect to time ( $t$ ) gives us $2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$ .
Remember, $x$ and $y$ are functions of time, so we use the chain rule!

We can simplify by dividing both sides of the equation by 2, giving us $x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0$ .
This makes the numbers easier to work with!

We're given $\frac{dx}{dt} = 8$ (the rate at which the bottom slides away) and $x = 2$ .
We need to find $\frac{dy}{dt}$ (how fast the top is sliding down).

When $x = 2$ , we can use $x^2 + y^2 = 289$ to find $y$ .
So, $2^2 + y^2 = 289$ , meaning $4 + y^2 = 289$ , so $y^2 = 285$ .
Since $y$ represents a length, we take the positive square root: $y = \sqrt{285}$ .

Calculating the fraction gives us $\frac{dy}{dt} \approx -0.953$ .
Since the question asks us not to include units, and to round to three decimal places if approximating, this is our final answer.

The top of the ladder is sliding down the wall at approximately $-0.953$ ft/s.
The negative sign indicates that the top is moving downwards.