# Need help deriving an expression for dy/dt

• Blake_ap1
In summary, the conversation is about finding the appropriate expression to use in Logger Pro to add a new calculated column for a vertical velocity vs time graph. The lab involves a pendulum and a Go! Motion sensor recording horizontal displacement and velocity. The horizontal velocity and position were calculated using specific expressions, but the appropriate expression for finding the vertical velocity is unknown. The conversation also includes discussions about differentiating displacement equations with respect to time and the use of the chain rule.
Blake_ap1

## Homework Statement

I need help finding the appropriate expression to use to add a "new calculated column" to produce a vertical velocity vs time graph in Logger Pro. The lab consisted of a ball tied to a rope, creating a pendulum. The Go! Motion sensor was set up to record the oscillations. The sensor recorded horizontal displacement and horizontal velocity. The x position vs time was calculated using the expression x = "R" - Ravg. The horizontal velocity was calculated using the expression y = L - sqrt(L^2-"x"^2).

I do not know what expression to use to find the vertical velocity.

R avg = (Rmax + Rmin) / 2 = 1.476 m. The section reads.

The horizontal velocity (vx) of the oscillating object was automatically calculated by the program. To find the vertical velocity (yy), create a new calculated column using the equation derivative("y"). The derivative (or dy/dt) is the rate of change of the vertical position with time. You should be able now to calculate the magnitude (v) of the resultant velocity. What is the appropriate equation for this calculation? Display vx, vy and v together on the same graph and include it in your lab report. Comment on the relationship between the three graphs.

Can someone point me in the right direction? Attatched are releative photos and the lab manual. I am doing Lab 4, and Working on the vertical velocity vs time section on page 25
Relevant equations The attempt at a solution
I'm having trouble Deriving the expression for dy/dt. I've hit a deadlock. I know I have to change things to the manner of y and θ. I know y = L - Lcosθ & x = Lsinθ

I know my expression needs to be along the lines of vy = (LxVx)/ (sqrt(x2+(L-y)2) + Ly)

#### Attachments

• Lab 4 1.png
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• Picture1.png
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Blake_ap1 said:
The horizontal velocity was calculated using the expression y = L - sqrt(L^2-"x"^2).
Looks more like a formula for y. It has the name and the dimension of the y in the first drawing.

Blake_ap1 said:
using the equation derivative("y")
Is that an expression your Logger Pro recognizes ? What IS the expression used for the horizontal velocity ?

Can you differentiate ##x^2 + (L-y)^2 = L^2 ## with respect to time ? ( I mean the analytical formula, not the LogPro expression)

BvU said:
Is that an expression your Logger Pro recognizes ? What IS the expression used for the horizontal velocity ?

Can you differentiate ##x^2 + (L-y)^2 = L^2 ## with respect to time ? ( I mean the analytical formula, not the LogPro expression)

Sorry, Y position was established using L-sqrt(L^2 - x^2)
Horizontal velocity and position were established with a motion sensor.

I derive the y position formula to obtain the velocity with respect to x. I’m not sure how to derive y with respect to time for dy/dt.
With respect to x I get vy = x/(sqrt(L^2 – x^2))

Blake_ap1 said:
With respect to x I get vy =
Differentiating one displacement wrt another will not give a velocity. That is dimensionally wrong. It might give you a gradient (and will in this case).

Blake_ap1 said:
I’m not sure how to derive y with respect to time for dy/dt.
Make an attempt. Use the chain rule.

## 1. What is the basic concept behind deriving an expression for dy/dt?

The basic concept behind deriving an expression for dy/dt is to find the rate of change of a dependent variable (y) with respect to an independent variable (t). This rate of change is represented by the derivative, dy/dt.

## 2. Why is it important to derive an expression for dy/dt?

Deriving an expression for dy/dt allows us to understand how the dependent variable is changing over time and how it is affected by the independent variable. This is crucial in many scientific fields, such as physics, chemistry, and biology, where understanding rates of change is essential.

## 3. What are the steps involved in deriving an expression for dy/dt?

The first step is to identify the dependent variable (y) and the independent variable (t). Then, use the appropriate rules of differentiation, such as the power rule or chain rule, to find the derivative. Finally, simplify the expression and substitute in any given values to get the final expression for dy/dt.

## 4. Can you provide an example of deriving an expression for dy/dt?

Sure, let's say we have the equation y = 2t^2 + 5t. To find dy/dt, we first identify y as the dependent variable and t as the independent variable. Then, we use the power rule to differentiate: dy/dt = 4t + 5. This is the final expression for dy/dt.

## 5. Are there any common mistakes to avoid when deriving an expression for dy/dt?

One common mistake is forgetting to include the derivative notation, dy/dt, in the final expression. It is also important to carefully apply the rules of differentiation and simplify the expression correctly. Additionally, make sure to check for any potential errors such as missing terms or incorrect substitution of values.

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