# 2 Related Rates Questions (Calc I)

• hell0
In summary, it appears that if y(0) = y(1) = 1, then the ladder will slide down, but if y(0) != y(1) then the ladder will rotate around its feet and fall over backwards.
hell0
Moved from a technical forum, so homework template missing
1. Suppose we have ladder laying against a wall, with x sub 0 = 1 and y sub 0 = 1. Given that y(1) = 1 and x′(1) = 1, find y'(1).

Okay so using Pythagorean's Theorem x^2 + y^2 = L, I found that the remaining side is sqrt(2). After taking the derivative I got 2x(t)x'(t) + 2y(t)y'(t) = 0. The only thing I'm confused about here is which values to plug in. I keep thinking that there's missing information in the problem or that there's some kind of typo. I've done questions like this one before, but the working of this specific one is so confusing to me.

2. Suppose we are draining an ice cream cone of the completely melted ice cream from a very small hole. Given that h(t) = 4 − 3t, r0 = 3 and V(t) = π/3 * (3/4)^2 * [h(t)]^3 , find V′(1).

This one I'm just stuck on completely :(.

edit: After taking the derivative and plugging in appropriately, i got -27π/16 for this one.

Last edited:
Hello Hell0,

Do I understand the center of the ladder is the point (x,y) and the initial condition is that the center is at (1,1) at t=0 ?
And then what ? It slides down ? Not if y(1) = y(0) = 1 !
Or does it rotate around its feet and fall over backwards in such a way that x'(1) = 1 when y(1) is y(0) = 1 again ?

BvU said:
Hello Hell0,

Do I understand the center of the ladder is the point (x,y) and the initial condition is that the center is at (1,1) at t=0 ?
And then what ? It slides down ? Not if y(1) = y(0) = 1 !
Or does it rotate around its feet and fall over backwards in such a way that x'(1) = 1 when y(1) is y(0) = 1 again ?
Good question. Honestly that's all the info I have. Nice professor I have, huh? Anyway I made the sides of the triangle be 1 and 1, and found the hypotenuse to be sqrt(2). Then used the theorem again but this time found x(1) via [x(1)]^2+(1/2)^2=sqrt(2)^2 since y(1) is given in the problem and found that x(1) is sqrt(7/4) or sqrt(7)/2. Then after taking the derivative of the Pythagorean Theorem I get 2x(t)x'(t)+2y(t)y'(t)=0. Plugging in I get 2(sqrt(7)/2)(1)+2(1/2)(y'(1))=0 and find y'(1) to be -sqrt(7)?

hell0 said:
y sub 0 = 1. Given that y(1) = 1
Assuming that means y(0)=1 and y(1)=1, as BvU notes, that is not consistent with the ladder sliding down. Please check you have those data correct.
It is unclear whether the x and y represent the coordinates of the ladder's midpoint or the variable ordinates of its endpoints, but since that just makes a factor of 2 difference everywhere it might not matter.

## What are related rates questions in Calculus I?

Related rates questions in Calculus I involve finding the rate of change of one variable with respect to another variable, when the two variables are related through a given equation. These types of questions often involve multiple variables and require the use of implicit differentiation.

## How do I solve related rates questions in Calculus I?

To solve related rates questions in Calculus I, you will need to follow a few steps: 1. Read the question carefully and identify the variables and their rates of change.2. Write an equation that relates the variables given in the question.3. Differentiate both sides of the equation with respect to time.4. Substitute in the known values and solve for the unknown rate of change.

## What are some common examples of related rates questions in Calculus I?

Some common examples of related rates questions in Calculus I include finding the rate at which the distance between two moving objects is changing, the rate at which the volume of a shape is changing, and the rate at which the area of a shape is changing.

## What are some tips for solving related rates questions in Calculus I?

Here are a few tips for solving related rates questions in Calculus I:1. Draw a diagram to help visualize the problem.2. Use the given information to write an equation that relates the variables.3. Use implicit differentiation to find the derivative of the equation.4. Substitute in the known values and solve for the unknown rate of change.

## Why are related rates questions important in Calculus I?

Related rates questions in Calculus I are important because they help develop the fundamental concept of derivatives and their applications. They also require students to think critically and apply their understanding of derivatives to real-world problems.

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