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Numerade Educator



Problem 22 Hard Difficulty

Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water depth at low tide is about 2.0 m and at high tide it is about 12.0 m. The natural period of oscillation is a little more than 12 hours and on June 30, 2009, high tide occurred at $ 6:45 AM. $ This helps explain the following model for the water depth $ D $ (in meters) as a function of the time $ t $ (in hours after midnight) on that day:

$ D(t) = 7 + 5 \cos [0.503(t - 6.75)] $
How fast was the tide rising (or falling) at the following times?

(a) 3:00 $ AM $
(b) 6:00 $ AM $
(c) 9:00 $ AM $
(d) Noon


a) $D^{\prime}(3)=2.39 \frac{\text { meters }}{\text {hour}}$
b) 0.926 $\mathrm{m} / \mathrm{hr}$
c) $D^{\prime}(9)=-2.28 \frac{\text {meters}}{\text {hour}}$
d) 1.21 $\mathrm{m} / \mathrm{hr}$

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Video Transcript

this problem is about high tides, and we have this equation D of tea, which represents the depth of the water and tea representative, represents the number of hours since midnight, and we're asked for how fast the tide is rising or falling at different times. And when you see the phrase how fast you want to think about rate of change and when you think about rate of change, that means derivative in calculus. And so we want to start by finding the derivative of the function so deep prime of tea, the derivative of 70 And then we move on to the other term and we can leave the five and take the derivative of co sign, which is negative sign. So we have negative sign of the entire inside. And then, according to the chain rule, we're going to multiply by the derivative of the inside. And so we would be multiplying by 0.503 Okay, so now that we have that derivative, we could simplify it if we wanted to. But chances are we're just going to be using a calculator from this point. So the calculators not going to mind if you don't simplify it for part A. We want to find the speed of the tide changing how fast the tide is rising or falling at 3 a.m. And 3 a.m. is going to be three hours since midnight, so we're going to use T equals three. We're going to substitute that into the derivative, use a calculator and you get approximately 2.39 meters per hour for part B. Three more hours have gone by, and now it's 6 a.m. So T equals six. We're going to substitute six into the derivative, use a calculator. Make sure your calculators and radiance, by the way, and we get 60.9 to 6 meters per hour. And for part C, it's now 9 a.m. So we're going to use T equals nine. Substitute that into the derivative. Put it in your calculator on you get approximately negative 2.28 Now, why is it negative? If the derivative is negative, that means you're quantity is decreasing. So that's telling us that the tide is going down and then finally, for party, it's noon, so T is 12 12 hours since midnight, and the rate of change AT T equals 12 is approximately negative. 1.21 Again, the tide must still be going down at this point in time.

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