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Fitting of second degree and exponential curves. practical. Fitting of second degree curve Prob.-. Fit the second degree curve for the above data. Solution- Let the second degree curve be y=a+bx+cx 2 ---------------------(1)

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Fitting of second degree and exponential curves.

### practical

Fitting of second degree curve

• Prob.-

Fit the second degree curve for the above data.

Solution-

Let the second degree curve be

y=a+bx+cx2 ---------------------(1)

By changing variable x to u=x-x , eq. (1) becomes

y=a'+b'u+c'u2 ---------------------(2)

Claim-To find a' ,b' & c'.

The normal eq.s to find a' ,b' & c' are

∑y=na'+b'∑u+c'∑u2 -------------(3)

∑uy= a'∑u+b'∑u2+c'∑u3 -----------(4)

∑u2y= a'∑u2+b'∑u3+c'∑u4 ----------(5)

Obtain the values of ∑y, ∑uy, ∑u2y ,∑u , ∑u2 , ∑u3 & ∑u4 . Use these values in eq. (3),(4) & (5) to get

a' ,b' & c'.

Put the values of u, a' ,b' & c‘ in eq. (2) , we get the required second degree curve.

a‘=31.9143 ,b' =5.3, c' =0.6429

Here x = 3

Let u=x-x =x-3

2) Fitting of exponential curve y=a bX

Prob.-

Fit the exponential curve y=a bX for the above data.

Solution-

Let the exponential curve be

y=a bX ---------------------(1)

Taking log (to the base 10) of both sides.

log (y)=log(a)+xlog(b)

V=A+BX ,Where log (y)=v, log(a)=A, log(b)=B

By changing variable x to u=(x-x)/10 we get

v=A +Bu-----------------------(2)

Then eq.(1) becomes

y=a bu ---------------------(3)

The normal eq.s to find A & B are

∑v=n A+ B ∑u -------------(4)

∑uv= A∑u+B∑u2-----------(5)

Obtain the values of ∑v, ∑uv, ∑u , ∑u2 .

Use these values in eq. (4),(5) to get A & B.

Taking antilog of A & B we get a & b.

putting u, a & b values in eq. (3) we get required eq.

A=2.3105,B=0.08295

a=204.41,b=1.2105

Here x = 1971

Let u=(x-x )/10=(x-1971)/10

3) Fitting of exponential curve y=a ebx

Prob.-

Fit the exponential curve y=a ebx for the above data.

Solution-

Let the exponential curve be

y=a ebx ---------------------(1)

Taking log (to the base 10) of both sides.

log (y)=log(a)+bxlog(e)

u=A+BX --------------------------(2)

Where log (y)=u, log(a)=A,

blog(e)=b x 0.4343=B

The normal eq.s to find A & B are

∑u=n A+ B ∑x -------------(3)

∑ux= A∑x+B∑x2-----------(4)

Use eq. 3 & 4 ,find the values of A & B .

Then

a= antilog (A), b=B/0.4343

Put these values of a & b in eq. 1 we get required curve.

A= -0.25347, B= 0.4617

a= 0.5578, b= 1.063

Prob.-Fit second degree & the curve y=a ebx ,y=a bX

Prob.-Fit second degree & the curve y=a bX

Prob.-Fit second degree curve & curve y=a ebx ,y=a bX