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.