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Suppose we're given the position \$x\$ , velocity \$\dot{x}\$ , and acceleration \$\ddot{x}\$ at \$t_1 \$.

We need to approximate the position \$x\$ at \$t_2\$.

In below picture, I understand how \$x_1\$ represents the vertical black length, and how \$ \color{blue}{\dot{x_1}(t_1)}h\$ represents the vertical blue length. However I don't get how \$\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}\$ represents the green length. I'm wondering if there a simple derivation for proving that the green vertical length equals \$\color{green}{\ddot{x}(t_1)}\dfrac{h^2}{2}\$ . Any ideas ? Thanks ! enter image description here

AgentS
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  • This question is from signals and systems, but I completely understand if this question belongs in math and I'll gladly post it in math if you think so.. (I got so used to electronics page that I didn't realize this till after finishing composing the q..) – AgentS Jun 15 '18 at 07:40
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    I think this is indeed better suited to Math SE. I'm usually not too strict on this type of stuff, but I believe the connection to electronics is too weak for this question ;-) – Sven B Jun 15 '18 at 08:14
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    Integral of h is.......? – Andy aka Jun 15 '18 at 08:44
  • @Andyaka Integral of h is h^2/2. That quadratic is what we get by integrating x''(t) = c. I now see clearly how the taylor quadratic looks geometrically. Thank you so much :) – AgentS Jun 15 '18 at 09:09
  • Is $\dot{x}$ derivative of x ? – Long Pham Jun 15 '18 at 09:21
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    Yes @LongPham if I recall correctly, physicists use $\dot{x}$ to represent derivative of $x$ with respect to time, and mathematicians use $x'$... Not sure about engineers hmm – AgentS Jun 15 '18 at 09:59

1 Answers1

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It is the contribution of the constant acceleration a.

After h time, the speed will have increased by ah. So the average extra speed over h will be ah/2. The extra distance due to that average extra speed will then be h*(ah)/2.

Which is the same as

$$ ah^2/2 $$

apalopohapa
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  • Ah so $ah^2/2$ is $a(t_2-t_1)^2/2$, and this term is part of the whole quadratic $x(t_1) + x'(t_1)(t_2-t_1) + x''(t_1)(t_2-t_1)^2/2$ . I see... that average acceleration interpretation is clever! Can we interpret something similar for the fourth term too ? $ah^3/6$ – AgentS Jun 15 '18 at 09:10
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    Yeah, it's called jerk – Vladimir Cravero Jun 15 '18 at 10:21