Solve Related Rates: More Homework Questions

In summary, the conversation discusses a problem involving a rope tied to a box and a truck, with one end of the rope passing over a pulley. The first part of the problem involves finding the speed at which the box is rising when the truck is 3 m from the point directly below the pulley. The second part involves finding the distance the truck needs to move to raise the box from the floor to a height of 2 m. The conversation includes equations and an attempt at solving the problem, with the summary concluding with a potential solution for part a and a suggested approach for part b.
  • #1
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Homework Statement



2) One end of a rope is tied to a box. The other end is passed over a pulley 5 m above the floor and tied at a level 1 m above the floor to the back of a truck. The rope is L meters long. If the rope is taut and the truck moves at ½ m/s:
a. How fast is the box rising when the truck is 3 m from point directly below the pulley?
b. How far will the truck have to move to raise the box from the floor to a height of 2 m?

Homework Equations



Known:
dx/dt = 0.5 m/s
x^2 + 16 = z^2 (length of rope from pulley to truck)
z + (5-y) = L (length of rope)

The Attempt at a Solution



for a:
2x dx/dt = 2z dz/dt

and dz/dt - dy/dt = 0, therefore dz/dt = dy/dt

dy/dt = x/z dx/dt
z = sqr (x^2 + 16)

dy/dt = x/[sqr (x^2 + 16)]* 0.5, where x=3

dy/dt = 0.3 m/s <---- someone verify?

For part B: Not sure, just know that I need to find x, giving y.
 
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  • #2


Hi there! Your attempt at solving part a looks correct. I would just add that you can also use the Pythagorean theorem to find x (length of the rope from the pulley to the box) when the truck is 3 m from the point directly below the pulley. It would look something like this:

x^2 + (5-y)^2 = L^2
x^2 + (5-1)^2 = L^2 (since the height of the truck is 1 m above the floor)
x^2 + 16 = L^2
x = sqr (L^2 - 16)

So when the truck is 3 m from the point directly below the pulley, x = sqr (L^2 - 16) = sqr (L^2 - 4^2).

For part b, you can use the same equation above, but this time you know the final height (2 m) and you need to find the distance the truck needs to move (x). So the equation would look like this:

x^2 + (5-y)^2 = L^2
x^2 + (5-2)^2 = L^2 (since the final height is 2 m)
x^2 + 9 = L^2
x = sqr (L^2 - 9)

Hope this helps! Let me know if you have any other questions.
 

FAQ: Solve Related Rates: More Homework Questions

What is the concept of related rates in mathematics?

Related rates is a mathematical concept that involves finding the rate of change of one variable with respect to another variable, where both variables are related by an equation.

How do you approach solving related rates problems?

To solve related rates problems, you need to identify the given and unknown variables, set up an equation that relates these variables, differentiate the equation with respect to time, substitute in the given values, and solve for the unknown rate of change.

Can you provide an example of a related rates problem?

An example of a related rates problem is a ladder leaning against a wall. As the base of the ladder moves away from the wall, the top of the ladder also moves. This creates a triangle, where the ladder is the hypotenuse and the base and height of the triangle are changing. The related rates problem is to find the rate at which the angle between the ladder and the ground is changing.

What are some common mistakes to avoid when solving related rates problems?

Some common mistakes to avoid when solving related rates problems include forgetting to differentiate the equation, using incorrect units, and not properly identifying and labeling the given and unknown variables.

How can related rates be applied in real-life situations?

Related rates can be applied in various real-life situations, such as calculating the rate at which a balloon is rising, the rate at which water is draining from a tank, and the rate at which the shadow of an object is changing. It can also be used in fields such as physics, engineering, and economics.

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