Answer:
[tex]\dfrac{d\theta}{dt}=-0.005 rad/s[/tex]
Step-by-step explanation:
We have in the image a representation of the triangle that is formed by the kite. The data that we have is:
[tex]\dfrac{dx}{dt}=2ft/s[/tex]
We need to find [tex]\dfrac{d\theta}{dt}=?[/tex].
As we can see in the attached image we can write the tangent of theta as:
[tex]tan\theta=\dfrac{100}{x}[/tex]
Then the cot of theta is:
[tex]cot\theta=\dfrac{x}{100}[/tex]
Derivating we have:
[tex]-csc^{2}\theta\dfrac{d\theta}{dt}=\dfrac{1}{100}\dfrac{dx}{dt}[/tex]
Changing [tex]csc^{2}\theta[/tex] we have:
[tex]-\dfrac{1}{sin^{2}\theta}\dfrac{d\theta}{dt}=\dfrac{1}{100}\dfrac{dx}{dt}[/tex]
as we can see in the figure attached [tex]sin\theta=\dfrac{100}{200}=\dfrac{1}{2}=0.5[/tex] using this we have:
[tex]-\dfrac{1}{(0.5)^{2}}\dfrac{d\theta}{dt}=\dfrac{1}{100}\dfrac{dx}{dt}[/tex]
Finally we just need to find [tex]\dfrac{d\theta}{dt}[/tex] then:
[tex]-\dfrac{1}{(0.5)^{2}}\dfrac{d\theta}{dt}=\dfrac{1}{100}\dfrac{dx}{dt}\\\\\dfrac{d\theta}{dt}=-0.0025\dfrac{dx}{dt}\\[/tex]
and we know that [tex]\dfrac{dx}{dt}=2ft/s[/tex] then:
[tex]\dfrac{d\theta}{dt}=-0.0025\times 2 rad/s\\\\\dfrac{d\theta}{dt}=-0.005 rad/s[/tex]
The rate at which the angle (in radians) between the string and the horizontal is decreasing when 200 ft of string have been let out is;
(dθ/dt) = -0.005 rad/s
We are given;
Height of kite above the ground = 100 ft
Horizontal speed; dx/dt = 2 ft/s
We want to find out the rate in which the angle in radians between the string and the horizontal decreasing when 200 ft of string have been let out.
Thus, i have attached an image showing the triangle formed by the given values.
From the image, we can see that the length of string is 200 ft and the height of kite is 100 ft.
Let x be the length of the horizontal and as such, we can say from the triangle that;
tan θ = 100/x
since we want to find the derivative at which the angle is decreasing, we will differentiate both sides to get;
sec²θ(dθ/dt) = (-100/x²)(dx/dt)
Now, from the triangle; x/200 = cos θ
1/ cos θ = 200/x
From trigonometric ratios, we know that; sec θ = 1/cos θ
Thus; sec θ = 200/x;
sec²θ = 200²/x²
Thus;
plugging the relevant values into sec²θ(dθ/dt) = (-100/x²)(dx/dt) gives;
(200²/x²)(dθ/dt) = (-100/x²)(2)
⇒ (40000/x²)(dθ/dt) = -200/x²
⇒ (dθ/dt) = -200/40000
⇒ (dθ/dt) = -0.005 rad/s
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