The depth of water in a leaking storage tank is d cm at time t hours after midnight on Sunday. The value of 𝑑 is given by 𝒅=𝟏𝟎−𝟐√𝟑𝒄𝒐𝒔𝟒𝒕𝒐−𝟐𝒔𝒊𝒏𝟒𝒕𝒐. Find
(i) The least and greatest depth of water possible with this model
(ii) The time at which the depth-first reaches these values
I wonder what 𝒕𝒐 means
I will assume that
𝒅=𝟏𝟎−𝟐√𝟑𝒄𝒐𝒔𝟒𝒕𝒐−𝟐𝒔𝒊𝒏𝟒𝒕𝒐
means
d = 10 - (2 sqrt 3) cos( 4 t ) - 2 sin( 4 t)
d d/dt = 8 sqrt 3 (sin 4t) - 8 cos 4 t
d d/dt = 0 at max or min
sin 4 t / cos 4 t= 1/sqrt3 = tan 4 t ( 1 sqrt 3 2 right triangle)
sin 4 t = 1/2 and cos 4t = sqrt 3 / 2 at max or min
your turn
d
I don't know what 𝒄𝒐𝒔𝟒𝒕𝒐 is, but If the function is
d = 10 - 2√3 cos4θ - 2sin4θ
then d = 10 - 4cos(4θ - π/6)
so the amplitude is 4, making the depth vary between 6 and 14
Now, cos(x) is a max when x=0 -- that makes d a min
cos(x) is a min when x = π -- that maxes d a max
Not sure what the 4to means in your equation, since you defined t to be in hours, I will read your equation as
d = 10 - 2√3cos(4t) - 2sin(4t)
dd/dt = -2√3(-sin(4t))(4) - 2cos(4t)(4)
= 8√3sin(4t) - 8cos(4t)
= 0 for a max/min of d
√3sin(4t) = cos(4t)
sin(4t) / cos(4t) = 1/√3
tan(4t) = 1/√3
4t = π/6 or 4t = π + π/6
t = π/24 or t = 7π/24 , where t is in hours
if t = π/24 , d = 10 - 2√3cos(π/6) - 2sin(π/6) = 10-3-1 = 6 cm
if t = 7π/24, d = .... = 10+3+1 = 14 cm
π/24 hours = appr 7.85 minutes
making the time of min at 12:07.85 am
To find the least and greatest depth of water possible with this model and the time at which the depth-first reaches these values, we need to understand and analyze the equation 𝑑 = 10 − 2√(3𝑐𝑜𝑠(4𝑡𝑜) − 2𝑠𝑖𝑛(4𝑡𝑜)).
(i) The least and greatest depth of water possible:
To find the least and greatest values of 𝑑, we can examine the equation and see that the depth will be minimum when the term inside the square root (√) is at its maximum, and maximum when the term inside the square root (√) is at its minimum.
Let's analyze the term inside the square root: 3𝑐𝑜𝑠(4𝑡𝑜) − 2𝑠𝑖𝑛(4𝑡𝑜).
Since cosine (𝑐𝑜𝑠) ranges from -1 to 1, and sine (𝑠𝑖𝑛) also ranges from -1 to 1, we can say that -1 ≤ 3𝑐𝑜𝑠(4𝑡𝑜) ≤ 1 and -1 ≤ 2𝑠𝑖𝑛(4𝑡𝑜) ≤ 1.
Now, let's substitute these values back into our equation:
d = 10 − 2√(3cos(4t) - 2sin(4t))
When the term inside the square root (√) is at its maximum (i.e. 3cos(4t) = 1 and 2sin(4t) = -1), the depth will be minimum. On the other hand, when the term inside the square root (√) is at its minimum (i.e. 3cos(4t) = -1 and 2sin(4t) = 1), the depth will be maximum.
Solving these equations, we get:
For minimum depth:
3cos(4t) = 1
cos(4t) = 1/3
4t = arccos(1/3)
t = arccos(1/3) / 4
For maximum depth:
3cos(4t) = -1
cos(4t) = -1/3
4t = arccos(-1/3)
t = arccos(-1/3) / 4
Hence, we have found the time at which the depth-first reaches its minimum and maximum values.
(ii) The time at which the depth-first reaches these values:
The time at which the depth-first reaches the minimum depth is t = arccos(1/3) / 4 hours.
The time at which the depth-first reaches the maximum depth is t = arccos(-1/3) / 4 hours.
Remember to convert the radian measure to hours if necessary.
Note: The exact numerical values of the minimum and maximum depths and the corresponding times would depend on the specific values of t and the initial conditions given in the problem.