Differential problem solving

In summary, the problem involves a balloon rising at a constant speed of 5ft/s and a boy cycling at a speed of 15 ft/s. The distance between the boy and the balloon is initially 45 ft. To find the rate of change of this distance 3 seconds later, we can use the equation 2s(ds/dt)+ 2x(dx/dt)+ 2y(dy/dt), where dx/dt and dy/dt are known and x, y, and s can be calculated using the given information and the distance formula s^2 = x^2 + y^2.
  • #1
venom_h
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0

Homework Statement



A balloon is rising at a constant speed of 5ft/s. A boy cycling along a straight road at a speed of 15 ft/s. When he passes directly under the balloon, it is 45 ft above him. how fast is the distance between the boy and the balloon increasing 3 seconds later?

Homework Equations



s^2= x^2+y^2
dx/dt = 15ft/s; dy/dt = 5ft/s

The Attempt at a Solution



2s(ds/dt)=2x(dx/dt)+2y(dy/dt)
Now, the problem is, how do i find the general equation for me to pluck in the t?
 
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  • #2
venom_h said:

Homework Statement



A balloon is rising at a constant speed of 5ft/s. A boy cycling along a straight road at a speed of 15 ft/s. When he passes directly under the balloon, it is 45 ft above him. how fast is the distance between the boy and the balloon increasing 3 seconds later?

Homework Equations



s^2= x^2+y^2
dx/dt = 15ft/s; dy/dt = 5ft/s

The Attempt at a Solution



2s(ds/dt)=2x(dx/dt)+2y(dy/dt)
Now, the problem is, how do i find the general equation for me to pluck in the t?
What do you mean by "the general equation"? You've already used the most general equation here- s^2= x^2+ y^2. You have, correctly 2s(ds/dt)+ 2x(dx/dt)+ 2y(dy/dt) (you might divide through by 2 to simplify.) You know dx/dt= 15 ft/s, dy/dt= 5 ft/s and certainly should be able to find x and y "3 seconds later" (how far does the boy go in 3 sec.? How far does the balloon rise in 3 sec.? And you can calculate s at that time. That's all you need.
 

What is differential problem solving?

Differential problem solving is a method used in scientific research to solve complex problems by breaking them down into smaller, more manageable parts. It involves analyzing the differences or changes between two or more situations or variables to better understand the problem at hand.

How is differential problem solving different from traditional problem solving methods?

Traditional problem solving methods typically involve finding a single solution to a problem. In contrast, differential problem solving focuses on understanding the underlying causes and differences between different situations or variables to find the best course of action.

What are some real-world examples of differential problem solving?

Differential problem solving can be applied in various fields, such as biology, economics, and engineering. For example, in biology, it can be used to understand the differences in gene expression between healthy and diseased cells. In economics, it can be used to analyze the differences in economic policies between countries. In engineering, it can be used to identify the differences in performance between different materials.

What skills are necessary for effective differential problem solving?

To be an effective differential problem solver, one must have strong analytical and critical thinking skills. It is also important to have a deep understanding of the subject matter and the ability to recognize patterns and differences between different variables or situations. Good communication and teamwork skills are also important for collaborating with others to solve complex problems.

How can differential problem solving benefit scientific research?

Differential problem solving can help researchers gain a deeper understanding of complex problems and identify potential solutions that may not have been apparent using traditional problem solving methods. It also allows for a more systematic and thorough analysis of data, leading to more accurate and reliable conclusions. Additionally, it encourages collaboration and interdisciplinary approaches, which can lead to innovative solutions to challenging problems.

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