# Thread: Find location using Taylor poly

1. ## Find location using Taylor poly

A car is at location f(0) = 10 miles with a velocity of f'(0) = 10 mi/min, acceleration f''(0) = 2 mi/min^2 and f'''(0) = -1 mi/min^3. Predict the location of the car at time t = 2 min.

Here is what I think needs to be done. A Taylor polynomial needs to be made using the above info. Once that is done plug in the time value and I should get the location.

Now how to interpet the info? All the functions are given with 0 as the input. So does that mean the start time of 0 or is this a Maclaurin Polynomial? I know that the dirivatives of f(x) form the coeffients of the powers of x in a Taylor/Maclaurin Polynomial.

The second dirivative of the location function is acceration but i do not know what the third dirivative is. the question says f'''(0) = -1 mi/min^3, i just don't see how this relates to finding location.

So I think the answer is something like this

P(x) = 10 + 10x + (2/2!)x + (2/2!)x^2 -(1/3!)x^3
P(2) = 32.67

i did some adding wrong but fixed it.

2. Hi

$\displaystyle f(x) = f(0+x) = f(0) + f'(0) \:x + \frac{f"(0)}{2}\:x^2 + \frac{f^{(3)}(0)}{6}\:x^3$

Spoiler:

$\displaystyle f(2) = 10 + 10 \times 2 + \frac{2}{2}\times 2^2 - \frac{1}{6}\times 2^3 = 32.67 m$

3. ok that looks like what I did. why did you choose to pick that formula?

4. This is Taylor-Maclaurin formula
It is the most appropriate one since you know all the derivatives values at t=0