rework final

course Mth 173

I am still a little unsure about this.

If your rate of depth change goes from 3.3 to 8.6, what is the average rate of depth change? &&&& 3.2654 + 8.6308 = 11.8962/2 = 5.9481

Once you know the average rate of depth change for this interval, what can you then find? Hint: What is the definition of the average rate of change of depth with respect to clock time?

&&&& The average rate of change of depth with respect to clock time over an interval is change in depth/change in clock time.

I promised to tell you the answers if anything was missing, but I can't resist the temptation to let you put the last key thing into place yourself:

You just determined that the average rate of depth change would be 5.9481.

You stated that 'The average rate of change of depth with respect to clock time over an interval is change in depth/change in clock time. '

You have the information you need to find the change in clock time, which you have already used in other calculations.

So how can you put all this together and get the change in depth?

The average rate of depth change is 5.9481

The change in clock time is 27.8-13.9=13.9

5.9481/13.9 = .4279 ?????

I am a little unsure about this..

You stated that 'The average rate of change of depth with respect to clock time over an interval is change in depth/change in clock time. '

Your last calculuation divides the average rate of depth change by the change in clock time. However your to find average rate of depth change your statement says to divide change in depth by change in clock time. That's very different.

If ave rate = `dy / `dt, then how do you find `dy when you know ave rate and `dt?

I'll put the answer below, but think through the question before you view the answer.

If ave rate = `dy / `dt, then algebraic rearrangement tells you that `dy = ave rate * `dt. So the change `dy in depth would be 5.948 * 13.9.

Go back and be sure you understand all the details of this problem. This will be very helpful to you in the course.