Solving the Difference Between h(t + Δt) and h(t): Explained

  • Thread starter Husaaved
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In summary, the reason why dividing both sides by Δt allows for the statement to be true is because it helps to rationalize the numerator and convert the difference into a sum, which makes it easier to solve and understand.
  • #1
Husaaved
19
1
h(t) = [itex]\sqrt{t - 1}[/itex]
h(t + Δt) = [itex]\sqrt{t + Δt - 1}[/itex]
h(t + Δt) - h(t) = [itex]\sqrt{t + Δt - 1}[/itex] - [itex]\sqrt{t - 1}[/itex]

So far so good. This is where I get confused:

[itex]\frac{h(t + Δt) - h(t)}{Δt}[/itex] = [itex]\frac{1}{\sqrt{t + Δt - 1} - \sqrt{t - 1}
}[/itex]

I don't understand why dividing both sides by Δt allows for this statement to be true. Can someone explain this to me? It would be very much appreciated.

Thanks a lot.
 
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  • #2
That should be
[itex]\frac{h(t + Δt) - h(t)}{Δt}[/itex] = [itex]\frac{1}{\sqrt{t + Δt - 1} + \sqrt{t - 1}
}[/itex]
to show it observe that
$$\frac{h(t + Δt) - h(t)}{Δt} = \frac{\sqrt{t + Δt - 1} - \sqrt{t - 1}
}{Δ t} \cdot \frac{\sqrt{t + Δt - 1} + \sqrt{t - 1}}{\sqrt{t + Δt - 1} + \sqrt{t - 1}} =\frac{1}{\sqrt{t + Δt - 1} + \sqrt{t - 1}
}$$
That is often called rationalizing the numerator, but in this case we don't really care that the numerator is rational we want to have a sum instead of a difference.
 
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  • #3
Thank you very much for your response, sorry for the typo.
 

1. What is the difference between h(t + Δt) and h(t)?

The difference between h(t + Δt) and h(t) is that h(t + Δt) represents the height of an object at a specific time t plus a small change (Δt) in time, while h(t) only represents the height at a specific time t.

2. Why is it important to understand the difference between h(t + Δt) and h(t)?

It is important to understand the difference between h(t + Δt) and h(t) because it allows us to make more accurate predictions about the behavior of objects over time. By considering small changes in time, we can better understand how an object's height or position changes over time.

3. How does solving the difference between h(t + Δt) and h(t) help in scientific research?

Solving the difference between h(t + Δt) and h(t) can help in scientific research by providing a more precise and accurate understanding of the behavior of objects. This can be especially useful in fields such as physics, engineering, and astronomy, where small changes in time can have a significant impact on the overall outcome.

4. Are there any real-world applications of understanding the difference between h(t + Δt) and h(t)?

Yes, there are many real-world applications of understanding the difference between h(t + Δt) and h(t). For example, in the field of robotics, understanding how an object's position changes over time is crucial for programming precise movements. It is also important in fields such as sports, where small changes in time can greatly affect an athlete's performance.

5. How can one solve the difference between h(t + Δt) and h(t) mathematically?

The difference between h(t + Δt) and h(t) can be solved mathematically by taking the derivative of the function h(t) with respect to time. This will give us the rate of change of h(t) over time, which is represented by h'(t). Then, we can use the formula h(t + Δt) = h(t) + h'(t)Δt to calculate the difference between h(t + Δt) and h(t) at any given time t.

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