V21

Thank you for your reply. What do you mean by inverting one of the variables? Does it mean that L = 1/f? Also, how do you find out this is the way to solve such kind of sums?

This question is mainly asking to rewrite a different graph equation (maybe parabolic, curved, etc) and changing it to a normal linear equation in the form y = mx+b. The question 5 and 6 can be seen in the picture, with further questions asking: i) Determine the dependent and independent variable. ii) Re-arrange the equation so that it is in the form of the straight-line equation. iii) What is the gradient and y-intercept of this re-arranged equation? iv) Sketch what the straight-line graph would look like. I know that linearisation works by first finding out what the y and x variables and then leaving x alone along with m multiplied to it. This works for simple examples like F = ma, where m is the gradient, a is x and F is y. When plotting on a graph, it will be F on the y axis and a on the x axis. However, for the fifth and sixth question, I'm not so sure how to do it. My first question is how to find out what is the independent and dependent variable in an equation with many variables and then how to re-arrange it to y = mx+b form. I thought for the fifth question the independent variable would be the length and the dependent variable the frequency. Then, frequency can be represented by 1/f. This would make f = 4/v * (l+e). This would give the gradient as 4/v I think and the x axis will be length, y axis will be f. I am not so sure. For the sixth question, I assumed v to be the dependent variable and u to be the independent. I made v = fu/u-f but I don't know whether any of these steps are correct and if they are, how to change it to the y = mx+ b format