Approximating instantaneous rate of change with average rate of change

the table gives a position s of a motorcyclist 40 between zero and three and cleaning including zero and three this is just saying that T is T is part of the interval or T in the interval between zero and three and we see that right over here where the distance traveled s is measured in meters and T is time in seconds assume the motorcyclist is accelerating during the three second period and they give us the information this is time between zero seconds and three seconds and here we have the corresponding distance that you could use a function of time the average velocity for T between one point five and two so T between one point five and two is 23 meters per second so what they did over here is they figured out well what is Delta s over delta T or in this interval and they figured out that it was 23 23 meters per second and you can verify that your change in s looks like it's 12.5 your change in time is point or actually this looks like it so let's see this is though now this is eleven point five actually let me see eleven point five your change in time is 0.5 eleven point five divided by 0.5 is twenty three so that's that makes sense and then they tell us the average velocity for T between two and two point five between two and two point five so change in our distance over change in time they say is 31 point eight meters per second and then they say estimate the instantaneous velocity at T equals two seconds and use this value to write the equation of a line tangent to s of T at the time T equals two so we can try to approximate we can approximate the slope of the tangent line right over here right when T equals two seconds by taking the average of the slopes of the tangent lines between one point five and two and two and two point five so essentially to approximate the slope of the tangent line we're going to take the average of these two rates of change right over here the average of these two slopes so let's do that so the average is going to be 23 plus 31 point 8 over 2 and let's see what is that equal to that is equal to 54 point 8 over 2 and what is that equal to let's see 54 divided by 2 is 27 so it's 27 point for 27 point four so we can use that as our approximation for the instantaneous rate of change for the slope of the tangent line and now we have to actually put we have to figure out what that equation actually is they don't just want the slope so this is the slope right over here and they say that they want it in point-slope form point-slope form and they remind us that T is the independent variable so when you're putting something in point-slope form it really just comes out of the definition of a line a line always has a constant slope so let's just imagine taking let's just imagine taking a a random point on that line T right this way T comma capital S a random point on the line on the tangent line here well the slope between that and this point is always going to be constant so what's the slope between this point and this point well what your change in s is going to be s minus 2 over your change in T which is T - t - oh sorry your s is s this is confusing sometimes s minus 30 point 2 over your change in t t minus 2 is equal to your slope or law anywhere along that tangent line you're going to have that slope 27.4 and then you multiply both sides by t minus 2 and you've put it in point-slope form so this is the same thing as s minus 30 point two is equal to twenty-seven point four times t minus 2