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3.3, #4, Given Problem.
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3.3, #4, Given Problem. |
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The maximum height will be reached when the velocity is equal to zero. Therefore, we find the derivative, |
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Part (a).
and set it equal to zero and solve. t = 2.5 s |
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Part (b).
We need to find out at what time the height is 96 feet, so we take the distance formula and set it equal to 96 and solve. We find that there are two times when this is true, namely at t = 3, and t = 2. |
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We begin with the velocity formula.
On its way up, at 96 ft. its velocity is 16 ft/s. On its way down, at 96 ft. its velocity is -16 ft/s. (The negative indicates that it is a velocity toward the earth) |
3.3, #16, Given Problem.
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3.3, #16, Given Problem. |
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Part (a)(i).
We calculate the average rate of change by calculating the slope over the interval. |
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Part (a)(ii).
We calculate the average rate of change by calculating the slope over the interval. |
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Part (a)(iii).
Again, we calculate the average rate of change by calculating the slope over the interval. |
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Part (b).
From our drawing we estimate the instantaneous rate of :  |
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We can tell this by looking at the above graph and seeing that as we move from left to right, the magnitude of the slope is getting closer to zero. |
3.3, #26, Given Problem.
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3.3, #26, Given Problem. |
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Part (a).
Since the sensitivity (S) is defined to be the rate of change of the reaction (R), we need to differentiate the R equation. |
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Here we finish the algebra and simplification. |
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Part (b).
We can see from this graph that when the strength of the stimulus (x) is small, the Reaction (red) is large and the sensitivity is large in value (although negative). When the stimulus is larger, the reaction is larger and the sensitivity is quite small. When the brightness is small, the sensitivity is at its highest When
the brightness is large the sensitivity is greatly reduced.
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