Evaluating the instantaneous rate of change using h = 0.1

In summary: Estimating" a rate of change by choosing a specific value of h is a very useful technique but it is not the same as evaluating the rate of change. In this particular case, with the function -(1/4)x^2+ 4x+ 12, the slope of the tangent to the graph at x= 1000 is -1/2 not -0.25. That is because the graph is not a straight line (it is a parabola) and the slope of a curve is not the same as the slope of the tangent.
  • #1
needingtoknow
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Homework Statement



Write a simplified algebraic expression, in terms of a and h, to estimate the instantaneous rate of change of the profit for each numeber of athletic shirts sold. Use h = 0.1.


The Attempt at a Solution



{ -0.25(1000+h)^2 + 4(1000+h) + 12 - [ -0.25(1000)^2 + 4(1000) + 12] } / h
{ -0.25(1000000+2000h + h^2) + 4000 + 4h + 12 - [-245 988] } / h
All the constants cancel out to zero which leaves me with
(-0.25h^2 - 496h)/h = -0.25h-496

The when I plug in h = 0.1. I get 496.025 when the answer in the back states 3.475. What am I doing wrong?
 
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  • #2
needingtoknow said:

Homework Statement



Write a simplified algebraic expression, in terms of a and h, to estimate the instantaneous rate of change of the profit for each numeber of athletic shirts sold. Use h = 0.1.


The Attempt at a Solution



{ -0.25(1000+h)^2 + 4(1000+h) + 12 - [ -0.25(1000)^2 + 4(1000) + 12] } / h
{ -0.25(1000000+2000h + h^2) + 4000 + 4h + 12 - [-245 988] } / h
All the constants cancel out to zero which leaves me with
(-0.25h^2 - 496h)/h = -0.25h-496

The when I plug in h = 0.1. I get 496.025 when the answer in the back states 3.475. What am I doing wrong?

Where does the 1000 come from? Presumably, it's from the number of shirts sold.

It would be helpful if you gave us all of the given information, including the profit function. It's possible that the variable in the profit function represents the number of shirts sold, in thousands.
 
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  • #3
If the profit for x shirts is given by ##- \frac{1}{4}x^2 + 4x + 12## then I don't see anything wrong. The rate of change could never be ~3.

Hmm, Mark44 makes a very good point, a larger secant could indeed give a lower rate of change.
 
  • #4
needingtoknow, the point is that you don't tell us what function you have or what the number of shirts is at which you are trying to find the derivative. Mark44 guessed that the number of shirts was 1000 and verty guessed that the function was -(1/4)x^2+ 4x+ 12. Please tell people things like that! Don't make them guess.
 
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  • #5
Oh sorry I missed to include that. The profit function is P(n) = -0.25n^2 + 4n + 12 where n is the number of athletic shirts sold, in thousands and P ios the profit in thousands of dollars.
 
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  • #6
and yes 1000 is the number of athletic shirts sold
 
  • #7
needingtoknow said:
Oh sorry I missed to include that. The profit function is P(n) = -0.25n^2 + 4n + 12 where n is the number of athletic shirts sold, in thousands and P ios the profit in thousands of dollars.

needingtoknow said:
and yes 1000 is the number of athletic shirts sold

Which means that, as I suspected, n should be 1 in your work, not 1000.

You need to work with this: { -0.25(1+h)^2 + 4(1+h) + 12 - [ -0.25(1)^2 + 4(1) + 12] } / h
 
  • #8
Oh yes that fixed it, thank you very much! I had a similar problem with an earlier question perhaps it is for the same reason. Thank you!
 
  • #9
By the way, I notice that you titled this "Evaluating the instantaneous rate of change using h= 0.1" but the problem actually asks you to estimate it. Those are very different things. You cannot, in general, "evaluate" a rate of change using a specific value of h- you have to take the limit as h goes to 0.
 
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FAQ: Evaluating the instantaneous rate of change using h = 0.1

What is the meaning of "evaluating the instantaneous rate of change using h = 0.1"?

When evaluating the instantaneous rate of change using h = 0.1, we are calculating the slope, or rate of change, of a function at a specific point. The value of h = 0.1 refers to the distance between the point and the nearby points used to calculate the slope.

Why is h = 0.1 commonly used when evaluating the instantaneous rate of change?

H = 0.1 is commonly used because it is a small enough interval to provide an accurate estimate of the instantaneous rate of change, but not so small that it becomes difficult to calculate or too sensitive to errors in measurement.

What is the process for evaluating the instantaneous rate of change using h = 0.1?

The process involves finding the slope of a function at a specific point by calculating the difference in the function's values at two nearby points that are h = 0.1 units apart. This value is then used to find the exact slope at the specific point.

What are the limitations of using h = 0.1 to evaluate the instantaneous rate of change?

One limitation is that h = 0.1 may not always provide an accurate estimate of the instantaneous rate of change for functions with sharp curves or steep slopes. It also may not be appropriate for functions with discontinuities or undefined points.

How can evaluating the instantaneous rate of change using h = 0.1 be applied in real-world situations?

This concept is commonly used in fields such as physics, engineering, and economics to analyze and predict changes in various quantities, such as velocity, acceleration, and cost. It can also be used to understand the behavior of natural phenomena, such as population growth or chemical reactions.

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