Something that does not make sense to me in the unit/subject of rate of change

harrison

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This question is in regards to grade 12 advanced function - specifically in the unit/subject of average rate of change and instantaneous rate of change.

Ok, so I am prepping for a final exam and the course I am taking offers practice exams with answers.

The question was:

A ball is tossed in the air on the planet Mars. It's height is modeled by h=-1.85(t-4)^2+31

a. Determine the maximum height of the ball

b. What velocity will the ball be travelling at when it hits the ground?

My answer:

a. 31

b.

-1.85(t-4)^2 +31 = 0

-1.85(t-4)^2 = -31

etc etc etc....

t=8.09

(8.09,0)

velocity = (-.0985 - .20416) / (8.1-8.08)

= -15.133

Therefore, the ball is moving downwards at 15.133 m/s.

Confirmed with first derivative

h=-1.85(t-4)^2+31

h' = -3.7(t-4)

= -3.7(8.09-4)

= -15.133


Hopefully you can see the picture as this is what their answer was. Makes no sense to me whatsoever what they did. Hopefully someone can explain....


Capture.jpg
 
It's just a different approximation.

You used the average velocity between 8.08 and 8.10 for your answer.

They used what might be called a right-hand approximation, simply the average velocity between 8.09 and 8.10. Too bad it was done poorly. This should have produced -15.15. The problem is the calculation for t = 8.09. The height is ASSUMED to be zero. That is incorrect. At t = 8.09, the height is 0.053 and the approximation is then -15.15.

Also, the attached mage is VERY SMALL. Feel free to improve that.

The exact answer is \(\displaystyle -\dfrac{\sqrt{5735}}{5}\) m/s, or around -15.146 m/s, so we're all just approximating. Far too much rounding going on. Maybe it's about Significant Figures, but still, far too much.
 
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The math they did in their last step is absolutely correct, but they screwed it all up by plugging in the wrong value. The answer that the height is 0 at t = 8.09 was only ever an approximation of the real answer, but they treated it as if were the 100% exact answer. Plugging it in ourselves, we see that:

\(\displaystyle -1.85(8.09-4)^2 + 31 =-1.85(4.09)^2 + 31 = -1.85(16.7281) + 31 = 0.053015\)

The exact answer for when h = 0 is an irrational number and is given by:

\(\displaystyle t = \sqrt{\dfrac{31}{1.85}} + 4\)

Using the correct values in their method produces (roughly) the same answer as you, certainly within tolerance for rounding errors:

\(\displaystyle \dfrac{-0.0985-0.053015}{8.10-8.09}=\dfrac{-0.151515}{0.01} = -15.1515\)
 
Thank you both for your answers!

From my understanding that although it is an estimate mine is a much closer approximation of the instantaneous velocity.

What worries me is that the answers are very far apart - 9.6 from practice test and 15.133 from my answer, possibly they will mark the answer as incorrect....
 
It should worry you that 9 is 40% less than 15. That's no good.

Don't decide that one method is ALWAYS better than another. In this case, your version was closer. In other cases, it will not be. There is no substitute for keeping your eyes open.
 
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