please answer the whole question or don\'t answer it at all. 5 stars Q1) Part a)

Question

please answer the whole question or don't answer it at all. 5 stars

Q1) Part a) Write out the linearized form of the following (show your working out) Part b) For each linear equation you found from part a, write down what you would plot on the x and y axis of a graph to produce a linear plot. Relate the Gradient and intercept of your linear graph to the constants A and B in the equations above. Part c) Can any equation be linearized? Does Linearising y-ax*2-bx+c help you find the values of a, b, c from the gradient and intercept of the linear plot?

Explanation / Answer

One can rewrite this equation as follows;

y-B = Ax2   ……. (1)

Now take the Log on both sides of (1) to get the following;

Log(y-B) = Log(A) + 2 Log(x)

Let us give relabels as follows;

Y = Log(y-B)

a = Log(A)

X = Log(x)

So that (2) can be written as ;

Y = a + 2 X.

This is the linearized form of given equation in (i).

To plot the linear plot, plot on X axis Log(x) values and on Y axis values of Log(y-B)

In the linearized form we can see that coordinated system is transformed to yield intercept on Y axis as a = Log(A) and slope is constant 2. The constant B in (i) is absorbed during the course of system transformation.

Consider equation (ii) y=BeAx

Let us call the equation

y=BeAx ……(2)

Low take the Ln (Natural log or log to the base e) on both sides of (2) to get the following;

Ln(y) = Ln(B) + A x

Let us give relabels as follows;

Y = Ln(y)

b = Ln(B)

Here we note that constant A and variable x is unchanged. The linearized form of (2) is then as follows;

Y = b + Ax

To plot the linear plot, plot on X axis x values and on Y axis values of Ln(y)

In the linearized form we can see that coordinated system is transformed to yield intercept on Y axis as b = Ln(B) and slope is constant A.

Every equation need not be linearized but with due approximation it can be.

Now consider the equation y = a x2-bx + c …. (3)

On assuming a 0, one can write (3) as follows;

y/a = x2 – bx/a + c/a

Y = x2 – Bx + C

Where Y = y/a, B = b/a and C = c/a

On squaring we can write

Y = (x- B/2)2 + C –B2/4

(Y – C + B2/4) = (x- B/2)2

Taking Log on both side we have

Log ((Y – C + B2/4)) = 2 Log((x- B/2))

Define

U = Log ((Y – C + B2/4)) and V = Log((x- B/2))

We can write

U = 2 V

This is the linearized form of (3)

In this example equation has three unknowns and it is rather difficult to obtain them from gradient and intercept only. It needs one more quantity.

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