Help me convert this relationship into a linear equation

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Discussion Overview

The discussion revolves around the challenge of converting a given equation into a linear relationship where 'a' represents the gradient of the line. Participants explore various methods to achieve this, including the potential use of logarithmic transformations and other mathematical techniques. The context includes both theoretical considerations and practical applications related to known values of 't' and 'h'.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses the need to convert an equation into a linear form with 'a' as the gradient, indicating difficulty in achieving this transformation.
  • Another participant suggests that while the relationship is not inherently linear, it may be possible to manipulate the equation using logarithmic techniques or other methods to derive a linear equation.
  • A further contribution emphasizes the importance of applying logarithms to both sides of an equation to create a linear relationship, providing an example of how this can be visualized on different axes.
  • Some participants inquire about alternative methods to calculate 'a' from the known values of 'h' and 't', seeking clarity on the approach.
  • One participant requests assistance in solving the equation for 'a', indicating ongoing confusion about the process.
  • Another response suggests a method involving algebraic manipulation, specifically subtracting a constant and squaring both sides to arrive at a quadratic equation in 'a'.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether a linear relationship can be derived from the original equation. Multiple competing views and methods are presented, with ongoing uncertainty about the best approach to take.

Contextual Notes

Limitations include the lack of clarity on the specific transformations applicable to the original equation, as well as the unresolved nature of the mathematical steps involved in deriving 'a'.

Who May Find This Useful

This discussion may be useful for individuals interested in mathematical transformations, particularly in the context of deriving linear relationships from non-linear equations, as well as those seeking to understand the manipulation of variables in physics or engineering applications.

kd001
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Is there anyway I can convert the attached equation into a linear relationship so that 'a' is the gradient of the straight line. 't' and 'h' are both variables. It doesn't matter exactly what the x and y values are. I just need 'a' to be the gradient of the straight line. I tried lots of different things but couldn't manage it. I would appreciate it very much if anybody can help me with this.

Thanks a lot.
 

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I typed the formula because I'm not sure if the attachment can be seen. I hope its clear enough

(at2)/2 = 0.25 - t(2ah)0.5
 
You can use the quadratic formula to get t as a function of h, but the relationship is not
linear.
 
I know the equation I wrote is not linear but is it possible to make it linear? For example sometimes it is possible to take logs to make a relationship linear. I'm not saying this is the case here but are there any other such 'tricks' to derive a linear equation from the one I wrote? I just need the gradient to be 'a'. I hope I'm making myself clear.

The reason I'm trying to do this is that I've got a series of known values for 't' and 'h' and I wanted calculate 'a' from the gradient of a straight line. Obviously if a linear relationship can be derived, the x and y values aren't going to be 't' or 'h' but some other value that can be given in terms of these variables.
 
Last edited:
If what I'm trying to do doesn't make sense to anyone, then what do you think would be the best way of calculating 'a', from the given values of 'h' and 't'?
 
When you took the logarithm of an equation such as y=ex to create a linear equation, you were supposed to be taking the logarithm of BOTH sides, so basically you have ln(y)=x. This means that you can have a linear equation (even though this expression is still the same as y=ex) if you plot the function on axes that are not linear. You can have the usual x axis, but you'll have to change the y-axis to ln(y).

i.e. at the point y=1, you should instead have ln(y)=0 on your axis. At y=e\approx 2.718 you should instead have y=1. At y=e2, y=2 etc.

OR you can keep the y-axis the same, but change the x-axis to ex so at the usual x=1 point, instead have x=e etc.

With a bit of ingenuity, you can apply the same concept to your problem.

However, I'm unsure if this is what you're looking for.
 
I am still struggling with this. Can someone at least show me how to solve the equation for 'a'?

Thanks a lot.
 
kd001 said:
Can someone at least show me how to solve the equation for 'a'?

Subtract 0.25 from both sides, then square both sides. You should end up with a quadratic equation in a.
 

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