Solving Derivatives Problems: Understanding dv/dt and dv/ds Differences

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In summary, The conversation is about a particle moving along a straight line and the relationship between its displacement, velocity, and acceleration. The question asks to show that a(t) = v(t) dv/ds and to explain the difference between dv/dt and dv/ds. The answer is that the former represents the change in velocity over time, while the latter represents the change in velocity over displacement. It is also mentioned that the chain rule is used to solve the first part of the question.
  • #1
Mathnewbie
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Hello can someone point in the right direction on this one.

A particle moves along a strainght line with displacement s(t), velovity v(t), and acceleration a(t). Show that

a(t) = v(t) dv/ds

Explain the difference between the meanings of the derivatives dv/dt and dv/ds.

Does dv/dt mean difference of velocity over the difference time ?

Does dv/ds mean difference of velocity over the difference displacement ?

Any help would be great? Thanks
 
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  • #2
Basically yes: how fast the speed changes "per foot" rather than "per second", for example.
To do the first part, use the chain rule:
[tex]\frac{dv}{dt}= \frac{dv}{ds}\frac{ds}{dt}[/tex]
 

FAQ: Solving Derivatives Problems: Understanding dv/dt and dv/ds Differences

What is the difference between dv/dt and dv/ds in derivatives?

Dv/dt and dv/ds both refer to the rate of change of a variable, but they represent different types of change. Dv/dt represents the rate of change over time, while dv/ds represents the rate of change with respect to a different variable. For example, if v is the velocity of an object, dv/dt would represent the acceleration of the object over time, while dv/ds would represent the acceleration of the object with respect to distance.

Why is it important to understand the difference between dv/dt and dv/ds in derivatives?

Understanding the difference between dv/dt and dv/ds is important because it allows us to accurately describe and calculate the behavior of a variable. If we use the wrong type of derivative, our calculations and predictions may be incorrect. Additionally, understanding the difference can help us better understand the relationship between variables and how they change over time or in relation to each other.

How do I know when to use dv/dt vs dv/ds in a derivatives problem?

In general, you will use dv/dt when the rate of change is being measured over time, such as in the case of velocity or acceleration. You will use dv/ds when the rate of change is being measured with respect to a different variable, such as in the case of acceleration with respect to distance. It is important to carefully consider the context of the problem to determine which type of derivative is appropriate.

Can you give an example of a problem that involves both dv/dt and dv/ds?

One example of a problem that involves both dv/dt and dv/ds is calculating the acceleration of a car as it travels around a circular track. In this case, dv/dt would represent the acceleration of the car over time, while dv/ds would represent the acceleration of the car with respect to the distance it has traveled around the track.

Are there any common mistakes to watch out for when solving derivatives problems using dv/dt and dv/ds?

One common mistake when using dv/dt and dv/ds is mixing up the variables and their units. It is important to keep track of which variable is being measured and ensure that all units are consistent. Another mistake is using the wrong type of derivative, so it is important to carefully consider the context of the problem to determine which type is appropriate.

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