Dave is going to leave academia and go into business building grain silos. A grain silo is a cylinder with a hemispherical top, used to store grain for farm animals. Here is a 3D view, a cross-section, and the top view.
If Dave is standing next to a silo of cross-sectional radius r = 11 feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12, 0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a. Enter your answer using interval notation. Round your answer to three decimal places.)
First, let's find the height of the cylinder, using the Pythagorean theorem. Since the radius of the cylinder is 11 ft, we have a right triangle with legs of length 11 and 12, and hypotenuse 11 + 12 = 23. Therefore, the height of the cylinder is $\sqrt{23^2 - 11^2} = 20$.
Now we know that silos have cross-sectional radius 11 ft and height 20 ft.
Let $(a,b)$ be the tangency point where line of sight #1 meets the cylinder. Then by the Pythagorean theorem, $a^2 + b^2 = 11^2$, so $b^2 = 121 - a^2$. Also, the slope of the line of sight is $\frac{b}{a + 12}$. Let's find the slope of the radial line through $(a,b)$, which has an equation in the form of $y=kx$. The radial line goes through the origin $(0,0)$ and the tangential point $(a,b)$, so we have $b = ka$ and $k = \frac{b}{a}$. Since the line of sight and the radial line are perpendicular, their slopes multiply to -1, so we have:
$\frac{b}{a+12} \cdot \frac{b}{a} = -1$
Substitute $b^2 = 121 - a^2$ into the equation:
$\frac{121 - a^2}{a(a+12)} = -1$
Now solve for $a$:
$121 - a^2 = -a^2 - 12a$
$12a = 121$
$a \approx 10.083$
Now we can find $b$:
$b^2 = 121 - a^2 \approx 121 - 100.667$
$b \approx 4.514$
To find the portion of the y-axis that Dave cannot see, we need to find the y-coordinates of the tangency points on the silo. For line of sight #1, the y-coordinate of the tangency point is $\approx 4.514$. For line of sight #2, we notice that the tangency point will have the same y-coordinate value, but in the negative direction, so it is $\approx -4.514$. Therefore, the portion of the y-axis that Dave cannot see is the interval:
$\boxed{[-4.514, 4.514]}$ (to three decimal places)
To find the portion of the y-axis that Dave cannot see, we need to determine the x-coordinate of the point where line of sight #1 is tangent to the silo.
Line of sight #1 is the line connecting Dave's eye level (point (12, 0)) to the tangent point on the silo.
Let's start by finding the slope of line of sight #1. The slope of this line can be calculated using two points: (12, 0) and the tangent point (a, h), where h is the height of the silo.
The slope of line of sight #1 is given by:
slope1 = (h - 0) / (a - 12) (1)
Now, let's find the slope of a radial line through the tangency point. Since this line is perpendicular to line of sight #1, its slope is the negative reciprocal of slope1.
The slope of the radial line is given by:
slope2 = -1 / slope1 (2)
Setting the calculations for slope1 and slope2 equal, we have:
(h - 0) / (a - 12) = -1 / slope1
Cross-multiplying, we get:
(h - 0) * slope1 = -1 * (a - 12)
Simplifying, we have:
slope1 * h = -a + 12
Now, let's substitute slope1 with its value from equation (1):
(h/(a - 12)) * h = -a + 12
Multiplying both sides by (a - 12), we get:
h^2 = -(a - 12)^2 (3)
We know that the silo has a spherical top, so the equation of a sphere is:
x^2 + y^2 = r^2 (4)
where r is the radius of the silo's cross-section.
Since the cross-sectional radius is given as 11, we can substitute r with 11 in equation (4):
x^2 + y^2 = 11^2
Simplifying, we have:
x^2 + y^2 = 121 (5)
In the top view of the silo, we can see that the tangent point lies on both equations (3) and (5).
Equating equations (3) and (5), we have:
-(a - 12)^2 = 121 - a^2
Expanding and simplifying, we get:
a^2 - 24a + 144 = 121 - a^2
Adding a^2 to both sides, we have:
2a^2 - 24a + 144 = 121
Subtracting 121 from both sides, we get:
2a^2 - 24a + 23 = 0
To solve this quadratic equation for a, we can use the quadratic formula:
a = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -24, and c = 23. Substituting these values into the quadratic formula, we have:
a = (-(-24) ± sqrt((-24)^2 - 4(2)(23))) / (2(2))
Simplifying, we get:
a = (24 ± sqrt(576 - 184)) / 4
a = (24 ± sqrt(392)) / 4
a = (24 ± sqrt(4 * 98)) / 4
a = (24 ± 2sqrt(98)) / 4
Simplifying further, we have:
a = 6 ± sqrt(98)
Rounding the answer to three decimal places, we have:
a ≈ 6 ± 9.899
a ≈ -3.899, 15.899
Note that a represents the x-coordinate of the tangency point. Since Dave is standing next to the silo, only the positive value of a is valid. Hence, a = 15.899.
Now, for Dave's vision to be partially obstructed, the x-coordinate of the tangent point (a) should be within the range of the x-axis, which is [-12, 12]. Thus, Dave cannot see the portion of the y-axis outside this interval.
Therefore, the portion of the y-axis that Dave cannot see is [-12, -3.899) U (12, ∞).
To find the portion of the y-axis that Dave cannot see, we need to find the x-coordinate where line of sight #1 is tangent to the silo. Let's go step by step to solve this problem.
Step 1: Determine the equation for line of sight #1.
Line of sight #1 is a line that connects Dave's position (12, 0) to a point on the silo's circumference. Since it is tangent to the silo, the slope of line of sight #1 will be perpendicular to the radius (x-axis).
Let's call the x-coordinate where line of sight #1 is tangent "a". The equation of line of sight #1 can then be expressed as: y = mx + b, where m is the slope and b is the y-intercept.
Step 2: Calculate the slope of line of sight #1.
The slope of line of sight #1 can be found by determining the slope of the line connecting the tangent point (a, 0) to Dave's position (12, 0). The slope, in this case, can be calculated as:
m = (0 - 0) / (a - 12) = 0 / (a - 12) = 0
Since the slope of line of sight #1 is 0, the equation of line of sight #1 becomes: y = b.
Step 3: Calculate the slope of the line perpendicular to the radius.
Since line of sight #1 is perpendicular to the radius, the line connecting the tangent point (a, 0) to the center of the silo (0, 0) will be the radius line. The slope of the radius line is:
m_radius = (0 - 0) / (0 - a) = 0
Step 4: Equate the slope of line of sight #1 and the slope of the radius line.
Since both slopes are 0, we can equate them:
0 = m = m_radius
Solving this equation, we find that the x-coordinate where line of sight #1 is tangent to the silo is a = 0.
Step 5: Find the portion of the y-axis Dave cannot see.
Now that we know a = 0, we can find the portion of the y-axis that Dave cannot see. It corresponds to the portion of the silo's circumference that lies between the tangency point (a, 0) and the top of the hemisphere (0, r), where r is the radius of the silo.
The y-coordinate of the tangent point is y = mx + b = 0 * 0 + b = 0 + b = b.
The y-coordinate of the top of the hemisphere is y = r.
Therefore, the portion of the y-axis Dave cannot see is the segment between y = b and y = r.
In this case, b = 0, so the portion of the y-axis Dave cannot see is the segment between y = 0 and y = r.
Finally, we substitute the given radius r = 11 feet into the equation. The portion of the y-axis that Dave cannot see is [0, 11] (interval notation), representing the vertical range from y = 0 to y = 11 feet.