A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up at a rate of 7 feet per second, how fast is the distance from the observer to the balloon increasing when the balloon is 14 feet high

Respuesta :

Answer:

[tex]\dfrac{dz}{dt}=0.65\ ft/s[/tex]

Explanation:

Given that

x= 150 ft

[tex]\dfrac{dy}{dt}= 7\ ft/s[/tex]

y= 14 ft

From the diagram

[tex]z^2=x^2+y^2[/tex]

When ,x= 150 ft and y= 14 ft

[tex]z^2=150^2+14^2[/tex]

[tex]z=\sqrt{150^2+15^2}[/tex]

z=150.74 ft

[tex]z^2=x^2+y^2[/tex]

By differentiating with respect to time t

[tex]2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}[/tex]

[tex]z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}[/tex]

Here x is constant that is why

[tex]\dfrac{dx}{dt}=0[/tex]

[tex]z\dfrac{dz}{dt}= y\dfrac{dy}{dt}[/tex]

Now by putting the values in the above equation we get

[tex]150.74\times \dfrac{dz}{dt}=14\times 7[/tex]

[tex]\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s[/tex]

[tex]\dfrac{dz}{dt}=0.65\ ft/s[/tex]

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

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