A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 800 ft/s when it has risen 3000 ft. (Round your answers to three decimal places.)
(a) How fast is the distance from the television camera to the rocket changing at that moment?
ft/s
(b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?
rad/s
Answers
============================================
Explanation for part (a)
t = time in secondsx = horizontal distance from the camera to the launch pady = vertical distance from the launch pad to the rocket's locationz = distance from camera to the rocket at time tAll distances mentioned are in feet.
We'll have a right triangle which allows us to apply the pythagorean theorem. Refer to the diagram below.
a^2+b^2 = c^2
x^2+y^2 = z^2
Apply the derivative to both sides with respect to t. We'll use implicit differentiation and the chain rule.
[tex]x^2+y^2 = z^2\\\\\frac{d}{dt}[x^2+y^2] = \frac{d}{dt}[z^2]\\\\\frac{d}{dt}[x^2]+\frac{d}{dt}[y^2] = \frac{d}{dt}[z^2]\\\\2x*\frac{dx}{dt}+2y*\frac{dy}{dt}=2z*\frac{dz}{dt}\\\\x*\frac{dx}{dt}+y*\frac{dy}{dt}=z*\frac{dz}{dt}\\\\[/tex]
Now we'll plug in (x,y,z) = (4000,3000,5000). The x and y values are given. The z value is found by use of the pythagorean theorem. Ie, you solve 4000^2+3000^2 = z^2 to get z = 5000. Or you could note that this is a scaled copy of the 3-4-5 right triangle.
We know that dx/dt = 0 because the horizontal distance, the x distance, is not changing. The rocket is only changing in the y direction. Or you could say that the horizontal speed is zero.
The vertical speed is dy/dt = 800 ft/s and it's when y = 3000. It's likely that dy/dt isn't the same value through the rocket's journey; however, all we care about is the instant when y = 3000.
Let's plug all that in and isolate dz/dt
[tex]4000*0+3000*800=5000*\frac{dz}{dt}\\\\2,400,000=5000*\frac{dz}{dt}\\\\\frac{dz}{dt} = \frac{2,400,000}{5000}\\\\\frac{dz}{dt} = 480\\\\[/tex]
At the exact instant that the rocket is 3000 ft in the air, the distance between the camera and the rocket is changing by an instantaneous speed of 480 ft/s.
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Explanation for part (b)
Again, refer to the diagram below.
We have theta (symbol [tex]\theta[/tex]) as the angle of elevation. As the rocket's height increases, so does the angle theta.
We can tie together the opposite side y with the adjacent side x with the tangent function of this angle.
[tex]\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(\theta) = \frac{y}{x}[/tex]
Like before, we'll apply implicit differentiation. This time we'll use the quotient rule as well.
[tex]\tan(\theta) = \frac{y}{x}\\\\\frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left[\frac{y}{x}\right]\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}*x - y*\frac{dx}{dt}}{x^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{800*4000 - 3000*0}{4000^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{3,200,000}{16,000,000}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{32}{160}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\[/tex]
Let's take a brief detour. We'll return to this later. Recall earlier that [tex]\tan(\theta) = \frac{y}{x}\\\\[/tex]
If we plug in y = 3000 and x = 4000, then we end up with [tex]\tan(\theta) = \frac{3}{4}\\\\[/tex] which becomes [tex]\tan^2(\theta) = \frac{9}{16}[/tex]
Apply this trig identity
[tex]\sec^2(\theta) = 1 + \tan^2(\theta)[/tex]
and you should end up with [tex]\sec^2(\theta) = 1+\frac{9}{16} = \frac{25}{16}[/tex]
So we can now return to the equation we want to solve
[tex]\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{25}{16}*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{d\theta}{dt} = \frac{1}{5}*\frac{16}{25}\\\\\frac{d\theta}{dt} = \frac{16}{125}\\\\\frac{d\theta}{dt} = 0.128\\\\[/tex]
This means that at the instant the rocket is 3000 ft in the air, the angle of elevation theta is increasing by 0.128 radians per second.
This is approximately 7.334 degrees per second.
The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s
Let x represent the distance from the camera to the rocket and let h represent the height of the rocket.
a)
[tex]x^2=h^2+4000^2\\\\2x\frac{dx}{dt}=2h\frac{dh}{dt} \\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\At \ h=3000\ ft, \frac{dh}{dt}=800\ ft/s;\\x^2=3000^{2} +4000^2\\x=5000\\\\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\5000\frac{dx}{dt}=3000*800\\\\\frac{dx}{dt}=480\ ft/s[/tex]
b)
[tex]tan(\theta)=\frac{h}{4000} \\\\h=4000tan(\theta)\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt} \\\\\\At\ h=3000\ ft;\\\\tan\theta = \frac{3000}{4000}=\frac{3}{4} \\\\sec^2(\theta)=1+tan^2(\theta)=1+(\frac{3}{4})^2=\frac{25}{16} \\\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt}\\\\800=4000*\frac{25}{16}* \frac{d\theta}{dt}\\\\\frac{d\theta}{dt}=0.128\ rad/s[/tex]
Hence, The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s
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Answers
Answer:
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Answers
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Pre-Algebra
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Step 1: Define
Identify
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Answer:
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Answers
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First, find the value of one variable by rearranging a equation:
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Answers
Answer:
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$3
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1 = (1 dollar)/(7 rand)
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Answers
Answer:
x= -2
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Answer:
qualitative
Step-by-step explanation:
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Required:
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Answers
Answer:
The answer is "0.6368 and 0.773".
Step-by-step explanation:
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Step-by-step explanation:
geometric mean theorem :
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Answer:
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Step-by-step explanation:
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Answer:
$16186.20
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Assuming that P represents the initial amount, A represents the end amount, r represents the annual interest rate, m represents the amount of times compounded per year, and t represents the amount of years, we can write this as
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Answers
Answer:C
Step-by-step explanation:
A correlation of -0.97 is a strong negative correlation while a correlation of 0.10 would be a weak positive correlation
Answer:
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Step-by-step explanation:
PLEASE ANSWER!!!!!!!
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Answers
Answer:
The graph is given below.
Step-by-step explanation:
X intercept = - 2
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Answers
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Step-by-step explanation:
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• • •
The mathematics department of a college has 6 male professors, 12 female professors, 14 male teaching assistants, and 11 female teaching assistants. If a person is selected at random from the group, find the probability that the selected person is a teaching assistant or a female.
The probability is ___.
(Type an integer or a fraction. Simplify your answer.)
Answers
Answer:
37/43
Step-by-step explanation:
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A. 15
B. 24
C. 6
D. 13
Answers
Answer:
15
Step-by-step explanation:
The figure has total 15 faces, the correct option is A.
What is a Hexagon?A hexagon is a polygon with six sides.
A hexagonal pyramid has 8 faces
From (2 hexagonal base + 6 lateral surfaces)
A hexagonal prism has 7 faces
From ( A hexagonal base + 6 lateral faces)
Total faces the figure has is 8 +7 = 15
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What is the domain of the function
v=m***
O x2
O
O O xe3
o
X> 3
ASAP
Answers
Answer:
The function will be exist if and only if :
[tex] \frac{ - x + 3}{2} > 0 \\ = > - x + 3 > 0 \\ = > - x > - 3 \\ = > x < 3 \\ \\ \therefore \bf \: domain \: \: \green{x < 3}[/tex]
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Answers
Answer:
x=13.2
Step-by-step explanation:
cos(43)=x/18
x=18×cos(43)
x=13.2
Answered by GAUTHMATH
A vending machine dispenses coffee into a twelve ounce cup he amount of coffee dispensed into the cup is normally distributed with a standard deviation of 0.006 ounce. You can allow the cup to overfill 4â% of the time. What amount should you set as the mean amount of coffee to beâ dispensed?
Answers
Answer:
this mean amount of coffee to be dispensed would be 11.99, approximately 12
Step-by-step explanation:
first of all we have this information available to answer this question.
standard deviation σ = 0.006 ounces
prob(x > 12) = 0.04
we use this formular to find the mean
z = x - μ/σ
the value of the z score at 4% is equal to 1.7507
such that
[tex]1.7507 = \frac{12-u}{0.006}[/tex]
we cross multiply from this stage
1.7507*0.006 = 12-μ
0.0105042 = 12-μ
μ = 12 - 0.0105042
herefore, the mean amount μ = 11.99 this can be approximated to 12
If a seed is planted, it has a 90% chance of growing into a healthy plant.
If 6 seeds are planted, what is the probability that exactly 2 don't grow?
Answers
Answer:
[tex]\displaystyle\frac{19,683}{200,000}\text{ or }\approx 9.84\%[/tex]
Step-by-step explanation:
For each planted seed, there is a 90% chance that it grows into a healthy plant, which means that there is a [tex]100\%-90\%=10\%[/tex] chance it does not grow into a healthy plant.
Since we are planting 6 seeds, we want to choose 2 that do not grow and 4 that do grow:
[tex]\displaystyle \frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}[/tex]
However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):
[tex]\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15[/tex]
Therefore, we have:
[tex]\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%[/tex]
Answer:
[tex] {?}^{?} However, this is only one case that meets the conditions. We can choose any 2 out of the 6 seeds to be the ones that don't grow into a healthy plant, not just the first and second ones. Therefore, we need to multiply this by number of ways we can choose 2 things from 6 (6 choose 2):
\displaystyle \binom{6}{2}=\frac{6\cdot 5}{2!}=\frac{30}{2}=15(26)=2!6⋅5=230=15
Therefore, we have:
\begin{gathered}\displaystyle\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \binom{6}{2},\\\\P(\text{exactly 2 don't grow})=\frac{1}{10}\cdot \frac{1}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot 15,\\\\P(\text{exactly 2 don't grow})=\boxed{\frac{19,683}{200,000}}\approx 9.84\%\end{gathered}P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅(26),P(exactly 2 don’t grow)=101⋅101⋅109⋅109⋅109⋅109⋅15,P(exactly 2 don’t grow)=200,00019,683≈9.84%
[/tex]
