How Can You Use a TI-83 to Solve Derivative Problems in Calculus?

  • Thread starter Thread starter Put a quark in it
  • Start date Start date
  • Tags Tags
    Derivatives
Click For Summary
To solve derivative problems using a TI-83, first input the derivative function A'(t) into the calculator's Y= section and graph it, focusing on x values greater than 10. Adjust the viewing window to set Xmin at 10 and Xmax at 30 for better visibility. Use the 2nd CALC function to find the maximum by defining left and right bounds around the maximum point. For part b, the second derivative A''(t) can also be graphed similarly, but manual calculations may yield more precise results, especially for logarithmic functions. It's recommended to strengthen logarithmic skills for accuracy in both manual and calculator-based solutions.
Put a quark in it
Messages
6
Reaction score
0
Greetings,

I'm having a little trouble solving a problem on my calc. homework. I'm not very proficient with my ti-83, which is why I'm having a problem. The problem I'm working on has two parts...

"The aerobic rating of a person x years old is modeled by the function...

A(t)=110[ln(x)-2]/x for x >= 10

a) At what age is a person aerobic rating largest?"

I understand that I am to take the derivative of the function, and set it equal to zero in order to maximize it, which results in...

(x*110[1/x] - 110[ln(x)-2] (1))/x^2

which simplifies into A'(t) = (110 - 110[ln(x)-2])/x^2 = 0

I can work this out on paper, but I'm not too confident with my abillity to do logarithms (which this requires), so I was wondering if anyone could explain how to solve this on the ti-83.

The second part of the question reads...

"b) At what age is a person's aerobic rating decreasting most rapidly?"

I know that I am now supposed to take the derivative of the derivative...

A''(t) = (x^2[110(1/x)]-[110-110[ln(x)-2]]*2x)/x^4

This problem I can't work out on paper (I tried and failed :cry: many times). The tutor at school recommended I just do this on the calculator, but again, I don't know how to do that. When I plug the A'' equation into Y1 on my graphing calculator and graph it, according to the calculator y = 0 at 11, but I know that's not right. The correct answer is 33 (approximately), but I don't know how to get it. Any information would be greatly appreciated. Thanks
 
Physics news on Phys.org
Part b is gross, I refuse! (Sorry. Heck, the homework's probably due by now, anyway, right?)

To solve part (a) on a TI-83, plug A'(t) into the calculator (in the y= section), and graph it. Notice that you only care about values for x>10. I suggest that you change the viewscreen by hitting window (the second button in the top row) and setting Xmin=10 and Xmax=30 (I picked Xmax because you could see 20 units along the x-axis before, so why change that fact?). Hit graph again. Now hit 2nd CALC (above trace, the fourth button in the top row) and select 4:maximum. A blinking cursor will appear on your graph, and the calculator will ask "Left Bound?" Use the right and/or left arrow keys to move the cursor to the left of the apparent maximum and hit enter. Now the calculator asks "Right Bound?" This time, use the right/left arrow keys to move the cursor to the right of what appears to be the maximum. Hit enter. Now the calculator will ask "Guess?" Hit enter again. The calculator will process the information for a few seconds, and the cursor will move to the maximum and the x and y values of the maximum will appear in the bottom of the screen.

BUT. I would suggest brushing up on your logarithmic skills!:-p
Manually setting the derivative to zero is NOT that bad, and you end up with lnx=3. That means e^3=x. Notice that that's the same value that the calculator gives you for the x value of your maximum--to about four or five decimal places. (If it's not, you punched something into the machine wrong.) But it's nice not to be dependent on a (specific) calculator. Also, very often exams will require an exact answer (ie still with all of the log, sqrt, sin, etc. still visible on your answer sheet).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
2K