Growth Rate is given by f(t) = 0.3t^2 +0.6t + 0.5

Don509

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At time t = 0, a seed is planted. After t weeks, the height of the plant is given by f(t) = 0.3t 2 +0.6t + 0.5 inches. At what rate is the plant growing after 8 weeks?

Here's what I've done:
Step 1: substitute 8 for x giving me - 0.3(64) + 0.6(8) + 0.5
Step 2: simplify 19.2 + 4.8 + 0.5
Step 3: 24.5cm plant height
Step 4: 24.5cm / 8 weeks = 3.1cm/week.

The correct solution is 5.4 cm/wk.

What am I doing wrong?

 
At time t = 0, a seed is planted. After t weeks, the height of the plant is given by f(t) = 0.3t 2 +0.6t + 0.5 inches. At what rate is the plant growing after 8 weeks?

Here's what I've done:
Step 1: substitute 8 for x giving me - 0.3(64) + 0.6(8) + 0.5
Step 2: simplify 19.2 + 4.8 + 0.5
Step 3: 24.5cm plant height
Step 4: 24.5cm / 8 weeks = 3.1cm/week.

The correct solution is 5.4 cm/wk.

What am I doing wrong?


Have you been taught about the derivative of a function?
 
At time t = 0, a seed is planted. After t weeks, the height of the plant is given by f(t) = 0.3t2 +0.6t + 0.5 inches. At what rate is the plant growing after 8 weeks?

Here's what I've done:
Step 1: substitute 8 for x giving me - 0.3(64) + 0.6(8) + 0.5
Step 2: simplify 19.2 + 4.8 + 0.5
Step 3: 24.5cm plant height
Step 4: 24.5cm / 8 weeks = 3.1cm/week.

The correct solution is 5.4 cm/wk.

What am I doing wrong?
You've found the average rate of growth over the entire period. It appears that you're expected to find the rate of growth at the end of the eighth week, which is a very different process.

In another thread, you've indicated a familiarity with taking limits. I think you may be expected to take the limit, as h "goes to infinity", of:

. . . . .\(\displaystyle \dfrac{f(8\, +\, h)\, -\, f(8)}{h}\)

If you've seen anything like the above in your class, then definitely try applying that here. This (just like the derivative, which comes from this limit) should return the expected (correct) value. ;)
 
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