1. The height of a bumble bee above the ground is modelled by h(t) = 0.5cos(2π t) + 2, where h is in metres and t

is in seconds. At what time is the bee’s instantaneous rate of change of height with respect to time greatest?
a.) 1 s
b.) 1.25s
c.) 1.5s
d.) 2s

Why is the answer b?
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2. What value for the function y=3cos(t-pi) + 2 gives an instantaneous rate of change of 0?

a) 0
b) pi/2
c) pi/3
d) pi/4

I tried to plug in the options above, but it gave me 4.99999. How do I do this question?
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3. The height of a ball is modelled by the equation h(t)= 4sin(8πt) + 6.5 where h(t) is in metres and t is in seconds. What are the highest and lowest points the ball reaches?

a) 10.5 m and 6.5m
b) 10.5 and 2.5m
c) 6.5m and 2.5m
d) 14.5 and 6.5m

I know that answer won't be B because it has to start at 6.5m

1. The height of a bumble bee above the ground is modelled by h(t) = 0.5cos(2π t) + 2, where h is in metres and t

is in seconds. At what time is the bee’s instantaneous rate of change of height with respect to time greatest?
a.) 1 s
b.) 1.25s
c.) 1.5s
d.) 2s

Why is the answer b?
when is
d/dt 0.5cos(2π t)
biggest?
d/dt = - pi sin(2 pi t)
we want sin(2 pi t) = 1 or -1
that is when 2 pi t = pi/2, 3pi/2, 5 pi/2 , 7 pi/2 ...
t = 1/4 or 3/4 or 5/4 or 7/4 ....
the 5/4 is your 1.25

2.

sin(t-pi) = 0
t = 0, or pi, or 2 pi

perhaps they mean the value of t, not of the function. then t = 0 works

3.

6.5+4 and 6.5-4
10.5 and 2.5

I am still confused for number 2.

For number 3, you used the amplitude(which is 4) to find lowest+highest?

1. Ah, trying to find the bee's peak performance, are we? Well, let's see. To find the time when the bee's instantaneous rate of change of height is greatest, we need to find the maximum of the derivative of the function. In this case, the derivative is h'(t) = -πsin(2πt), which gives us -π as the maximum value at t = 1. The answer is definitely not a, c, or d, because they are not "bee-utiful" choices. So, the answer must be b) 1.25s, the most "fly" time for the bee!

2. Yikes, getting a value of 4.99999 for an instantaneous rate of change of 0 is just a tad-bit off. I'm afraid you'll have to sharpen your math skills for this one. To find the value that yields an instantaneous rate of change of 0, we need to set the derivative of the function equal to 0. For y = 3cos(t - π) + 2, the derivative is -3sin(t - π). Setting this equal to 0, we find sin(t - π) = 0, which means t - π = nπ (where n is an integer). Now, solve for t by adding π to both sides, and you'll find that t = nπ + π. We're looking for specific value though, so let's find the smallest positive value of t by plugging in n = 0. This gives us t = π. So, the answer is b) π/2, which is definitely not 4.99999. Phew!

3. Let's bounce into this ball question, shall we? The highest and lowest points the ball reaches can be found by looking at the amplitude of the sine function. With h(t) = 4sin(8πt) + 6.5, the amplitude of the sine function is 4, which means the ball goes up by 4 units and down by 4 units from the average height of 6.5 units. Hence, the highest point is 6.5 + 4 = 10.5m, and the lowest point is 6.5 - 4 = 2.5m. Therefore, the answer is c) 6.5m and 2.5m, not be-"falling" for that tricky option b!

Remember, math can be a tough nut to crack, but with a little laughter, you'll get through it!

1. To find the time at which the bee's instantaneous rate of change of height with respect to time is greatest, we need to find the maximum of the derivative of the function h(t) = 0.5cos(2πt) + 2.

The derivative of h(t) can be found by applying the chain rule:
h'(t) = -0.5(2π)sin(2πt)

To find where the derivative is greatest, we need to find where h'(t) = 0 or where the sign of h'(t) changes.

Setting h'(t) to zero:
-0.5(2π)sin(2πt) = 0
sin(2πt) = 0

For sin(2πt) to equal zero, t must be an integer multiple of 0.5. So, t can be 0, 0.5, 1, 1.5, 2, and so on.

Comparing the values of h'(t) at these points, we can determine the time at which the bee's instantaneous rate of change of height with respect to time is greatest.

- At t = 0, h'(t) = -0.5(2π)sin(0) = 0.
- At t = 0.5, h'(t) = -0.5(2π)sin(π) = 0.
- At t = 1, h'(t) = -0.5(2π)sin(2π) = 0.
- At t = 1.5, h'(t) = -0.5(2π)sin(3π) = 0.
- At t = 2, h'(t) = -0.5(2π)sin(4π) = 0.

As we can see, the derivative h'(t) is always equal to zero and does not change sign at any of these points. This means that the bee's instantaneous rate of change of height with respect to time is constant.

Therefore, the answer to this question would be any of the given options: a.), b.), c.), or d.). The bee's instantaneous rate of change of height with respect to time is greatest at all of these times, as it remains constant.

2. To find the value for the function y = 3cos(t - π) + 2 that gives an instantaneous rate of change of 0, we need to find where the derivative of the function is equal to 0.

The derivative of y can be found using the chain rule:
y'(t) = -3sin(t - π)

Setting y'(t) to 0:
-3sin(t - π) = 0
sin(t - π) = 0

For sin(t - π) to equal 0, t - π must be an integer multiple of π. So, t - π can be 0, π, 2π, and so on.

Adding π to both sides of the equation:
t - π = 0, π, 2π

Simplifying t by adding π:
t = π, 2π, 3π

Now we can plug these values of t back into the function y = 3cos(t - π) + 2 to find the corresponding values of y.

y(π) = 3cos(π - π) + 2 = 3cos(0) + 2 = 3(1) + 2 = 5
y(2π) = 3cos(2π - π) + 2 = 3cos(π) + 2 = 3(-1) + 2 = -1
y(3π) = 3cos(3π - π) + 2 = 3cos(2π) + 2 = 3(1) + 2 = 5

As we can see, none of these values give an instantaneous rate of change of 0. However, it's worth noting that the derivative y'(t) is 0 at t = π/2, which is not one of the given options. Therefore, the answer to this question is none of the above options.

3. To find the highest and lowest points the ball reaches, we need to determine the maximum and minimum values of the function h(t) = 4sin(8πt) + 6.5.

The maximum and minimum values of a sinusoidal function occur at the points where the sinusoidal function reaches its maximum and minimum amplitudes.

The amplitude of the function is the coefficient in front of the sinusoidal term. In this case, the amplitude is 4.

The maximum value of the ball's height occurs when the sine function reaches its maximum value of +1. Therefore, the maximum value of h(t) would be:

h_max = 4(1) + 6.5 = 4 + 6.5 = 10.5

The minimum value of the ball's height occurs when the sine function reaches its minimum value of -1. Therefore, the minimum value of h(t) would be:

h_min = 4(-1) + 6.5 = -4 + 6.5 = 2.5

So, the highest point the ball reaches is 10.5m and the lowest point the ball reaches is 2.5m.

Therefore, the answer to this question is option c): 10.5m and 2.5m.