Problem 7. The height of an object in motion is given by...

Problem 7. The height of an object in motion is given by $h(x) = \sqrt{1 + 2x}$ meters, where \\ $x$ is the number of seconds after it starts moving.\ (1) On average, how rapidly is the height of the object changing between $x = 1$ second \\ and $x = 1.1$ second?\\ Hint: This question is asking for the average velocity of the object.\ (2) How rapidly is the height changing at $x = 1$ second?\\ (3) Write the equation for the tangent line to the graph of $h(x)$ at $x = 1$.

Transcript

00:01 In this video, we are going to be going over the average rate of change, instantaneous rate of change, and we're also going to be touching upon some other terms like tangent lines, secret lines, and velocity.

00:12 So, we know that the average rate of change is simply the rate of change between two values of a function, and we can calculate this by finding the slope of the secret line between them.

00:22 We know that the instantaneous rate of change is the rate of change at one point on a function, point a, and we can calculate it by finding the limit as h approach is 0 of ffa plus h minus f of a over h.

00:35 And it's also important to note that the tangent line is the line that intersects the graph of the function f of x at one point and one point only.

00:43 And that the second line is the line between two points on the function f of x, and it only intersects it at those two points.

00:51 Finally, since we are going to be working with the velocity function in this question, we need to remember that the velocity function, v of x, is simply the derivative of the position function.

01:02 So in this question, we are going to be working with the position function, f of x is equal to x squared plus x, and what they ask us to do in part a is to find the average velocity for x changing from 1 to 3 seconds.

01:14 So we know that what they're asking us to find here is the average rate of change between when x is equal to 1 and x is equal to 3.

01:21 Or in simpler terms the slope of between the points 1f1 and 3f3 so if we quickly wanted to evaluate this we would get 1 2 and 312 so what part a wants us to do is to find the slope between these two points so we know that our slope is change in y over change in x so we have 12 minus 2 over 3 minus 1 which is a same thing as 10 over 2 or 5.

02:02 And since we know that in the question, they mentioned that y is in meters and x is in seconds, we know that our velocity is going to be 5 meters per second.

02:20 So in part b, they want us to find the average velocity or the average rate of change again, but in this case, instead of it being between 1f1 and 3f3, they want it between 1f1 and and a new point, 1 plus h, f of 1 plus h.

02:40 So we're going to evaluate this again, like we did with our previous points, and we will get 1 plus h, and if you want to find out what f of 1 plus h is really quick, you'll see that that's 1 plus h squared, plus 1 plus h, and that gives us h squared plus 3h, plus 2.

03:04 So we have the point 1 plus h, h squared plus 3h plus 2.

03:13 For part b, we know that they want us to find the slope between these two points, so we can plug them in.

03:23 Change in y, over change in x.

03:27 Then in the numerator, we're going to have h squared plus 3h plus 2 minus 2, all over 1 plus h minus 1...

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