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Math  /  Calculus

QuestionA rider's height, hh, in metres on a Ferris wheel is modelled by h(t)=−49cos⁡(2πt142)+52h(t)=-49 \cos \left(\frac{2 \pi t}{142}\right)+52 where tt is time in seconds. Solve for the AROC between t=138t=138 and t=140t=140. Does this represent when the rider is going to the top of the ride, or coming back towards the ground? How do you know? Since the slope is negative, the rider is getting closer to the ground Since the slope is negative, the rider is heading backwards through time Since the slope is negative, the rider is moving to the top of the ride Since the slope is less than 1 , the rider is heading back towards the ground Previous Page Next Page Page 11 of 12

Studdy Solution

STEP 1

What is this asking? We need to find how much the rider's height changes on average between 138\text{138} and 140\text{140} seconds, and figure out if they're going up or down! Watch out! Make sure to correctly plug the time values into the height formula and don't mix up the order when calculating the average rate of change!
Also, remember AROC is just the slope between two points!

STEP 2

1. Calculate h(138)h(138)
2. Calculate h(140)h(140)
3. Calculate AROC
4. Interpret the AROC

STEP 3

Let's **plug in** t=138t = 138 into our **height formula**: h(138)=−49⋅cos⁡(2π⋅138142)+52 h(138) = -49 \cdot \cos\left(\frac{2\pi \cdot 138}{142}\right) + 52

STEP 4

h(138)=−49⋅cos⁡(276π142)+52 h(138) = -49 \cdot \cos\left(\frac{276\pi}{142}\right) + 52 h(138)=−49⋅cos⁡(138π71)+52 h(138) = -49 \cdot \cos\left(\frac{138\pi}{71}\right) + 52

STEP 5

h(138)≈−49⋅cos⁡(6.152)+52 h(138) \approx -49 \cdot \cos(6.152) + 52 h(138)≈−49⋅0.963+52 h(138) \approx -49 \cdot 0.963 + 52 h(138)≈−47.192+52 h(138) \approx -47.192 + 52 h(138)≈4.808 h(138) \approx 4.808 So, at 138\text{138} seconds, the rider is at approximately 4.808\text{4.808} meters **high**!

STEP 6

Now, let's do the same thing for t=140t = 140: h(140)=−49⋅cos⁡(2π⋅140142)+52 h(140) = -49 \cdot \cos\left(\frac{2\pi \cdot 140}{142}\right) + 52

STEP 7

h(140)=−49⋅cos⁡(280π142)+52 h(140) = -49 \cdot \cos\left(\frac{280\pi}{142}\right) + 52 h(140)=−49⋅cos⁡(140π71)+52 h(140) = -49 \cdot \cos\left(\frac{140\pi}{71}\right) + 52

STEP 8

h(140)≈−49⋅cos⁡(6.22)+52 h(140) \approx -49 \cdot \cos(6.22) + 52 h(140)≈−49⋅0.995+52 h(140) \approx -49 \cdot 0.995 + 52 h(140)≈−48.762+52 h(140) \approx -48.762 + 52 h(140)≈3.238 h(140) \approx 3.238 At 140\text{140} seconds, the rider is at approximately 3.238\text{3.238} meters.

STEP 9

The **AROC** (average rate of change) is the change in height divided by the change in time: AROC=h(140)−h(138)140−138 \text{AROC} = \frac{h(140) - h(138)}{140 - 138}

STEP 10

AROC=3.238−4.808140−138 \text{AROC} = \frac{3.238 - 4.808}{140 - 138}

STEP 11

AROC=−1.572 \text{AROC} = \frac{-1.57}{2} AROC≈−0.785 \text{AROC} \approx -0.785

STEP 12

The **AROC** is *negative*, which means the rider's height is *decreasing*.
They're coming down!

STEP 13

The AROC of the rider's height between 138\text{138} and 140\text{140} seconds is approximately −0.785-0.785 meters per second.
Since it's negative, the rider is going down, closer to the ground.

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