How Fast is the Distance Changing as a Particle Moves Along y=sqrt(x)?

In summary: You are given that "the particle passes through the point (4,2)" and "its x-coordinate increases at a rate of 3 cm/s" so at that moment x= 4 and dx/dt= 3. That gives (1/2)(4)^{-1/2}(3)= (1/2)(3/2)= 3/4 cm/s.In summary, the particle is moving along the curve y=\sqrt{x} and at the point (4,2) its x-coordinate is increasing at a rate of 3 cm/s. To find how fast the distance from the particle to the origin is changing at this instant, we use the equation of the curve and the chain rule to
  • #1
bob1182006
492
1

Homework Statement


A particle is moving along the curve [itex]y=\sqrt{x}[/itex]. As the particle passes through the point (4,2), its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant?

Homework Equations


The Attempt at a Solution


I made a diagram of the curve, connected the point (4,2), x=4, and the origin by a right triangle with z being the hypotenuse, x = 4, and y = 2.

so [itex]z^2=x^2+y^2[/itex], after differentiating I arrive at
[tex]\frac{dz}{dt}=\frac{1}{z}(x\frac{dx}{dt}+\frac{dy}{dt})[/tex]
i know dx/dt 3 cm/s, but I have no idea how to find dy/dt, I have a feeling that I have to use the equation of the curve but I'm not sure at all
 
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  • #2
You're given the acceleration (3cm/s2) at the X-ordinates.

Maybe you should be intergrating to find the velocity.
 
  • #3
bob1182006 said:
I have a feeling that I have to use the equation of the curve but I'm not sure at all
The equation is given, and you haven't employed it yet
 
  • #4
Sorry I don't get what anyone is saying, so far the book hasn't taught integration so I don't think I should use it.

f(x) said:
The equation is given, and you haven't employed it yet

sry I was doing this in the middle of the night ><

so dy/dt = derivative of [itex]y=\sqrt{x}[/itex] with respect to t
but how would the chain rule work there? would it be like
[tex]\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}[/tex]
right? so then I'd arrive to
[tex]\frac{dy}{dt}=\frac{1}{2\sqrt{x}}\frac{dx}{dt}[/tex]
so I'd plug in x=4, dx/dt = 3 cm/s, and obtain dy/dt
ok i think I am getting this now thanks
 
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  • #5
Yes, exactly! You were told that the particle is moving along the curve [itex]y= \sqrt{x}= x^{1/2}[/itex] so, using the chain rule,
[tex]\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}= (1/2)x^{-1/2}\frac{dx}{dt}[/tex].
 

FAQ: How Fast is the Distance Changing as a Particle Moves Along y=sqrt(x)?

1. What are related rates in calculus?

Related rates in calculus refer to the rate at which one quantity changes in relation to another quantity. It involves finding the derivative of one variable with respect to another in order to determine how they are changing together.

2. How are related rates used in real-world applications?

Related rates are used in real-world applications to solve problems involving changing quantities, such as rates of change in physics, chemistry, and economics. They are also used in engineering and other fields to analyze and optimize systems.

3. What is a particle in related rates?

In related rates, a particle refers to a point in space that is moving and changing in relation to other particles or variables. It can represent a physical object, such as a car or a particle in a chemical reaction, or it can be a mathematical point used to represent a changing quantity.

4. How do you set up a related rates problem?

To set up a related rates problem, you first need to identify the changing quantities and determine how they are related. Then, use the given information and apply the chain rule to find the derivative of one variable with respect to another. Finally, set up an equation using the given information and solve for the unknown rate.

5. What is the chain rule in related rates?

The chain rule in related rates is used to find the derivative of one variable with respect to another when there is a function within a function. It states that the derivative of the outer function multiplied by the derivative of the inner function gives the derivative of the entire function.

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