How Do You Solve Problems on Tangent and Velocity for a Sliding Ladder?

In summary, the conversation discusses a problem involving a ladder leaning against a building and sliding down the wall. The location of the ladder is described by parametric equations and the average velocity is calculated for various time intervals. The problem also asks to find a time interval where the average velocity is -20 ft/sec. Finally, there is a discussion on instantaneous velocity and a picture of the graph of the function y(t) is provided for reference. The true/false questions are based on the concepts of average and instantaneous velocity.
  • #1
Soohyun
1
0
Hi guys! I had a question on the bolded sections below and wondered how to solve this in a step by step solution. I already have the answers, but don't know how to solve it. I will show the whole problem so that any content needed is available for sections e and d which I am stuck on. Thanks!

A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder at time t seconds is given by the parametric equations (7+2t,0).

View attachment 8753


(a) The location of the top of the ladder will be given by parametric equations (0,y(t)). The formula for y(t)=
√625−(7+2t)2

. (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.)

(b) The domain of t values for y(t) ranges from 0 to 9
(c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places):
time interval ave velocity time interval ave velocity
[0,2] -0.775 [2,4] -1.23
[6,8]. -3.23 [8,9] -9.80
(d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a= 8.75
(e) Using your work above and this picture of the graph of the function y(t) given below, answer these true/false questions: (Type in the word "True" or "False")

View attachment 8752


The top of the ladder is moving down the wall at a constant rate
TF F

The foot of the ladder is moving along the ground at a constant rate
TF Correct AnT

There is a time at which the average velocity of the top of the ladder on the time interval [a,9] is 1 ft/sec
TF Correct Answer: F
There is a time at which the average velocity of the top of the ladder on the time interval [a,9] is 0 ft/sec
TF Correct Answer:F
There is a time at which the average velocity of the top of the ladder on the time interval [a,9] is -100 ft/sec
TF Correct Answer:T
There is a time at which the average velocity of the top of the ladder on the time interval [a,9] is less than -100 ft/sec
TF Correct Answer: T
 

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(d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this time interval is -20 ft/sec i.e. a= 8.75

$\bar{v} = \dfrac{\Delta y}{\Delta t} = -20 = \dfrac{0-\sqrt{625-(7+2a)^2}}{9-a}$

$-20 = \dfrac{-\sqrt{4(144-7a-a^2)}}{9-a}$

$10 = \dfrac{\sqrt{(16+a)(9-a)}}{9-a}$

$10 = \sqrt{\dfrac{16+a}{9-a}}$

$a = \sqrt{\dfrac{884}{101}} \approx 8.75$

for part (e), you know $\bar{v} = \dfrac{\Delta y}{\Delta t}$, which is the slope of a secant line passing between any two distinct points on the curve.

Instantaneous velocity, $v = \dfrac{dy}{dt}$, is the slope of the line tangent to a single point on the curve.

Try answering the T/F questions and give reason(s) for your choice.
 

FAQ: How Do You Solve Problems on Tangent and Velocity for a Sliding Ladder?

1. What is the difference between a tangent and a velocity?

A tangent is a straight line that touches a curve at only one point, while velocity is the rate of change of an object's position over time.

2. How are tangents and velocity related?

Tangents and velocity are related in that the slope of a tangent line at a specific point on a curve is equal to the velocity of the object at that point.

3. What is the formula for calculating the slope of a tangent line?

The formula for calculating the slope of a tangent line is m = lim (Δy/Δx), where Δy represents the change in y-values and Δx represents the change in x-values.

4. How can tangents and velocity be used to solve real-world problems?

Tangents and velocity can be used to solve real-world problems by helping to determine the instantaneous rate of change of an object's position, which can be useful in predicting future motion and making decisions based on that information.

5. What are some common applications of tangents and velocity in science and engineering?

Tangents and velocity have various applications in science and engineering, such as in motion analysis, optimization problems, and designing systems that require precise control and accuracy.

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