The sun is setting at the rate of 1/4 deg/min, and appears to be dropping perpendicular to the horizon, as depicted in figure 6.2.5. How fast is the shadow of a 25 meter wall lengthening at the moment when the shadow is 50 meters long?
angle a = tan^-1 (.5)
da/dt = -.25 deg/min = -.009436 radians/min
x = 25 = y tan a
y = 25/tan a
dy/dt = -25 sec^2 a da/dt /tan^2 a
I can't see the diagram, but it appears that if the shadow is x, then
25/x = tanθ
so,
-25/x^2 dx/dt = sec^2θ dθ/dt
-25/2500 dx/dt = (1+1/4)(-1/4 * π/180)
Now just solve for dx/dt
Well, if the sun is setting at a rate of 1/4 deg/min, I bet it's saying goodbye to all the other celestial bodies on its way down. It's probably telling the moon, "See you later, alligator!" Then again, the moon might reply, "After a while, crocodile!" Ah, the sun and the moon, the original dad joke duo.
Now, let's get back to the serious stuff. Since we have a 25-meter wall, and its shadow is 50 meters long, we want to find the rate at which the shadow is lengthening. So, to solve this, we need to use a bit of trigonometry.
Since the sun is dropping perpendicular to the horizon, we have a right triangle formed by the wall, its shadow, and a line from the top of the wall to the sun.
To find the rate at which the shadow is lengthening, we'll need to use the tangent function. The tangent of an angle in a right triangle is equal to the ratio of the opposite side to the adjacent side.
In this case, the opposite side is the lengthening shadow, and the adjacent side is the height of the wall. So, we can say:
tan(theta) = (lengthening shadow) / (height of wall)
To find how fast the shadow is lengthening, we need to differentiate both sides of the equation with respect to time. So, let's assume d(theta)/dt is the rate at which the sun is setting. And since we know that the rate at which the sun is setting is 1/4 deg/min, we have:
d(theta)/dt = 1/4 deg/min
Now we can differentiate both sides of the equation:
sec^2(theta) (d(theta)/dt) = (d(lengthening shadow)/dt) / (height of wall)
We want to find (d(lengthening shadow)/dt) when the shadow is 50 meters long. So, we need to find theta when the shadow is 50 meters long.
Using the tangent function again, we have:
tan(theta) = (lengthening shadow) / (height of wall)
tan(theta) = 50 m / 25 m
tan(theta) = 2
Now we can find theta using the inverse tangent function:
theta = arctan(2)
Now substitute the value of theta into our equation for sec^2(theta) (d(theta)/dt):
sec^2(arctan(2)) * (1/4 deg/min) = (d(lengthening shadow)/dt) / 25 m
And there you have it! That's how you find the rate at which the shadow is lengthening when the shadow is 50 meters long.
But remember, this is all just math and calculations. Let's not forget to appreciate the beauty of a sunset and the fun of telling dad jokes along the way!
To find the rate at which the shadow of a wall is lengthening, we need to use similar triangles.
Let's assume the wall is vertical, and the shadow it casts is horizontal.
Let's say the length of the shadow at a given time is S meters, and the height of the sun above the horizon is H degrees.
Based on the given information, we know that the sun is setting at a rate of 1/4 degree per minute. Therefore, the change in angle of the sun with respect to time is dH/dt = -1/4 degrees per minute.
Using the basic trigonometric relationship between the height of the sun, the length of the shadow, and the wall height, we have:
tan(H) = wall height / shadow length
Differentiating both sides of this equation implicitly with respect to time, we get:
sec^2(H) * dH/dt = (dw/dt) / S
Here, dw/dt represents the rate at which the shadow length is changing with respect to time.
We are given that the shadow length is 50 meters when we want to find the rate at which it is changing. Therefore, S = 50 meters.
We need to find dw/dt when S = 50 meters, so we rearrange the equation to solve for dw/dt:
dw/dt = S * sec^2(H) * dH/dt
Since we want to find dw/dt when the shadow is 50 meters long, we need to find the value of H when S = 50 meters. We can use the trigonometric relationship above to solve for H:
tan(H) = wall height / shadow length
tan(H) = 25 meters / 50 meters
tan(H) = 1/2
Taking the inverse tangent of both sides of the equation, we find:
H = tan^(-1)(1/2)
Using a calculator, H is approximately 26.57 degrees.
Now we have all the values we need to find dw/dt:
S = 50 meters
H ≈ 26.57 degrees
dH/dt = -1/4 degrees per minute
Substituting these values into the equation, we get:
dw/dt = (50 meters) * sec^2(26.57 degrees) * (-1/4 degrees per minute)
Calculating the value on the right side of the equation, we get:
dw/dt ≈ -12.5 meters per minute
Therefore, the shadow of the 25-meter wall is lengthening at a rate of approximately -12.5 meters per minute when the shadow is 50 meters long. The negative sign indicates that the shadow is getting shorter as the sun sets.
To find out how fast the shadow of the wall is lengthening, we can use related rates. Let's denote the length of the shadow as "S" and the rate at which it is lengthening as "ds/dt". We are given that the sun is setting at a rate of 1/4 deg/min, which means the angle of depression of the sun is changing at this rate.
Now, let's consider the geometry of the situation. We have a right triangle formed by the wall, its shadow, and the sun. The length of the wall is 25 meters, and we are given that the length of the shadow is 50 meters. Let's denote the height of the wall as "h" and the angle of depression of the sun as "θ".
We can set up the following equation using trigonometry:
tan(θ) = h/S
To find ds/dt, we need to differentiate this equation with respect to time. Since h and θ are not changing, their derivatives will be zero:
0 = (d/dt)(tan(θ)) = sec^2(θ)(dθ/dt)
Now, we have an expression for dθ/dt in terms of the rate of change of the angle. We are given that dθ/dt = 1/4 deg/min.
Now, let's differentiate the original equation with respect to time:
D[h/S]/dt = D[tan(θ)]/dt
To find ds/dt, we need to find dS/dt. Rearranging the equation, we can solve for dS/dt:
dS/dt = (dS/dθ) * (dθ/dt)
Now, we need to find dS/dθ, which is the derivative of S with respect to θ. From the original equation, we can isolate S:
S = h/tan(θ)
Differentiating this equation with respect to θ:
dS/dθ = dh/dθ * (d/dθ)[1/tan(θ)]
Using the quotient rule and the derivative of tan(θ), we can simplify this equation:
dS/dθ = -h * sec^2(θ)
Therefore, we now have all the components to find ds/dt. Let's substitute the given values:
ds/dt = (dS/dθ) * (dθ/dt)
= (-h * sec^2(θ)) * (1/4 deg/min)
We need to find ds/dt when S = 50 meters. To find θ, we can use the trigonometric definition of tan(θ):
tan(θ) = h/S
tan(θ) = h/50
θ = arctan(h/50)
Now, substitute θ into the expression for ds/dt:
ds/dt = (-h * sec^2(θ)) * (1/4 deg/min)
= (-25 * sec^2(arctan(h/50))) * (1/4 deg/min)
Therefore, to find the rate at which the shadow of the wall is lengthening when it is 50 meters long, we substitute h = 25 and evaluate the expression for ds/dt.