Calculus

posted by .

Great! I understand now that
C0=f(a)
c1=f'(a)
c2=f"(a)/2

Now, If I am to find the parabolization of the equation x^2-x at x=2, then
c0=x^2-x=2^2-2=2
c1=2x-1=2(2)-1=3
c2=2/2=1

So, the equation (taken from c0+c1(x-a)+c2(x-a)^2) is >>
2+3(x-2)+1(x-2)^2??

Is this correct?

Thanks

In calc, we are studying parabolization. It is the linerazation of a parabola. The linerazation of a standard line is L(x) = b0+b1(x-a) when f(x) is at x=a

b0 = f(a)
b1=f'(a)

The parabolization of f(x) at x=a is given by the equation
P(x)=c0+c1(x-a)+c2(x-a)^2.

f(a) = P(a)
F'(a) = P'(a)
f''(a) = P''(a)

I need to find a formula for c0, c1, and c2 in terms of f(a), f'(a) and f"(a)

Im sure that I need to find the first and second derivitive of the equation c0+c1(x-a)+c2(x-a)^2. Im just not sure where to start...

Thanks!

Matt

Actually, linearization is approxiamtion by a straight line while Parabolization is approximation of a function by a parabola.

You start from:

f(a) = P(a)
f'(a) = P'(a)
f''(a) = P''(a)

Then you insert

P(x)=c0+c1(x-a)+c2(x-a)^2

in here.

So,

P(a) = C0
P'(x) = C1+2c2(x-a) --->
P'(a) = C1
P''(x) = 2C2 -->
P''(a) = 2C2

This means that:

f(a) = P(a) = C0
f'(a) = P'(a) = C1
f''(a) = P''(a) = 2C2

And it follows that:

P(x) = C0+C1(x-a)+c2(x-a)^2 =
f(a) + f'(a)(x-a) + f''(a)/2 (x-a)^2

There is an easy way to check. If you parabolize a parabola, you should get the same thing back. So, let's see:

2+3(x-2)+ (x-2)^2 =

2 + 3x-6 + (x^2 - 4x + 4) =

2-6+4 + 3x-4x + x^2 =

x^2 - x.

So, it's indeed the same!

Similar Questions

1. drwls, is this correct,math

drawls this is from the previous post is this correct. solve the system by subtraction. 5x-3y=13 4x-3y=11 equation#2 woudl be 4x-3y=11 -3y = -4x+11 y = (4)/(3) x - (11)/(3) so now i substitute it to equation #1 5x - 3((4)/(3)x - (11)/(3)= …
2. math

I need help with this question. Write the equation of the line which contains the point)0, -3) and whose slope is 4. A general line equation can be written as y = mx + c, where m is the slope and c a constant. so we can write this …
3. Calculus

I am unsure of how to take the derivative of this equation. It may be the exponents that are giving me trouble but I'm not sure exactly. Find the equation of the tangent line to the curve 4e^xy = 2x + y at point (0,4). On the left …
4. Math

2x represents that Clyde is twice (2 times) as old as Bonnie. 6 years from now, Clyde will be twice as old as his cousin Bonnie will be then. If Clyde is now 16 years old, how many years old is Bonnie?
5. Calculus

Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0. I know that normal of the latter equation is (3,2,6) but now what do I do?
6. Algebra 2

I have a question and I can not figure it out. Any answers to the following along with any explanation would be great! Here is the equation: x=.2y^2-2y-3 Now I know that is an equation for a parabola. Now, I need the following: Equation …
7. calculus - what is the derivative?

Let f(x) = 2x^2 + 1 a. Find the derivative f' of f. b. Find an equation ofthe tangent line to the curve at the point (1,3). c. Sketch the graph of f. *Can someone please help me?
8. Pre Cal.--Check my work?

Solve the system of equations: x - y - z = 2; x + 2y - 2z = 3; 3x - 2y - 4z = 5?
9. Calculus

Solve the differential equation dy/dx = -xe^y and determine the equation of the curve through P(1,2) I tried solving the differential equation and I get y = log(x^2/2 + C). Is this correct?
10. math2

I was solving a problem. In It r have been given that two lines are perpendicular to each other The equation of one line 2x-3y+4=0 Now i have to find the equation of another line We can find it by finding the slope of the abovementioned …

More Similar Questions