Answer :
Answer:
24 ft/sec
Step-by-step explanation:
Let the string be the hypotenuse of a right triangle with a constant feet of 50
The vertical leg is considered as the height of the kite
The horizontal leg of the kite is not given, but increases at speed of 26 ft/sec
The length of the leg should be called leg x , the length of the string L at any point is known by
L² = 50² + x².
The String release rate is given as dL/dt. That is the amount looking for presently when L = 150. When L = 130, x that is found by substitution
130² = 50² + x², then x = √(130² - 50²) = 120 ft
Applying differentiation method for change of L
2L dL/dt = 0 + 2x dx/dt
When L = 130 and then x = 120
2(130) dL/dt = 2(120) (26)
dL/dt = 24 ft/sec
The rate of change of the length of the string with time is given by
applying chain rule of differentiation.
When the kite is 130 ft. away, the child must let out the string at a rate of 24
ft./second.
Reasons:
The given information are;
Height of the kite = 50 ft.
Horizontal speed of the kite due to the wind = 26 ft./s
Required:
The rate at which the child should let the string out when the kite is 130 ft.
away.
The length of the string, s = √(x² + 50²)
The speed of the string = [tex]\dfrac{ds}{dt}[/tex]
By chain rule of differentiation, we have;
[tex]\dfrac{ds}{dt} = \mathbf{\dfrac{ds}{dx} \times \dfrac{dx}{dt}}[/tex]
Where;
[tex]\dfrac{dx}{dt}[/tex] = The horizontal speed of the kite = 26 ft./s
[tex]\dfrac{ds}{dx} = \dfrac{d\sqrt{x^2 + 50^2} }{dx} = \dfrac{x}{\sqrt{x^2 + 50^2}}[/tex]
Therefore;
[tex]The \ speed \ of \ the \ string = \dfrac{ds}{dt} = 26 \times \dfrac{x}{\sqrt{x^2 + 50^2}} = 26 \times \dfrac{x}{s}[/tex]
When s = 130, we get;
[tex]\dfrac{ds}{dt} = 26 \times \dfrac{\sqrt{130^2 - 50^2}}{130} \approx 24[/tex]
The speed of the string when the kite is 130 ft. away is 24 ft./sec.
The child must let out the string at a rate of 24 ft./sec.
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