Problem

Find the instantaneous rate of change for each fu…

02:39

Problem 16 Easy Difficulty

Find the instantaneous rate of change for each function at the given value.
$s(t)=-4$ $t^{2}-6$ at $t = 2$

Answer

-16

Related Courses

Calculus 1 / AB

Finite Mathematics and Calculus with Applications

Chapter 11

The Derivative

Section 3

Rates of Change

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Video Transcript

we want to find an instantaneous rate of change or the function s A T is it would turn negative 40 squared minus six. AT T is equal to two. So we have the formula right above. So it's written slightly differently since the formula they give us in the book is with respect f as opposed toe s and X and set of tea. But still, we'll work out the same. So in this case, are a is going to be too. So I like to do this in steps. So the first step I like to do it's to find what is not enough. But it would be s of a plus h So two plus page is so it's plug. Listen so we'd have negative or two plus h squared minus six. So now we can go ahead at expe ed back. So you haven't making before times four plus for H plus H squared by the six. Then we're going to have negative 16 by the 16 H minus four H square minus six that we could just simplify that down and that would give us negative 20 to minus 16 h minus or h squared then I like to go ahead and do s of two plus h minus s of to. So let's figure out what s up to is right here. So square to that would give us negative or times for minus six, which is going to be negative 16 minus six, which gives us negative 22. So plug in that it now. So we have negative 22 minus 16 h minus or H squared than minus negative 22. So these negatives here accounts out and then positive trying to a negative 22 counts out with each other. Then we can move on to los step of taking the limit as a trip purchase zero of us of two plus H minus s of to, and then this would be all over age. So now kind of as a pro tip. Once you get to this point here, if you're working with a polynomial for your original function and for some reason you can't evenly divide the h out, then you need to go back and look at your algebra again. But in this case knows how it can evenly divide each the things in the numerator. So we have the limit as H Purchase zero of negative 16 minus or each. Now that's just the line. So it's continuous, so we can just apply the limit directly. So it's gonna give us negative 16.0, which is negative. 16. So the instantaneous writer change at a busy go to term t 0 to 2 is negative 60.

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