Given:
Let x=34 miles
y=44 miles
[tex]\frac{dy}{dt}=41 miles/hr[/tex]
To find:
[tex]\frac{d\theta}{dt}[/tex]
Solution:
[tex]Hypotenuse=\sqrt{(base)^2+(Perpendicular\;side)^2}[/tex]
Using the formula
[tex]Hypotenuse=\sqrt{(34)^2+(44)^2}=55.6 miles[/tex]
[tex]sec\theta=\frac{Hypotenuse}{Base}[/tex]
Using the formula
[tex]sec\theta=\frac{55.6}{34}[/tex]
[tex]tan\theta=\frac{Perpendicular\;side}{Base}[/tex]
Using the formula
[tex]tan\theta=\frac{y}{34}[/tex]
Differentiate w.r.t t
[tex]sec^2\theta\frac{d\theta}{dt}=\frac{1}{34}\frac{dy}{dt}[/tex]
Using the formula
[tex]\frac{d(tanx)}{dx}=sec^2 x[/tex]
Substitute the values
[tex](\frac{55.6}{34})^2\times \frac{d\theta}{dt}==\frac{1}{34}\times 41[/tex]
[tex]\frac{d\theta}{dt}=\frac{41\times (34)^2}{34\times (55.6)^2}[/tex]
[tex]\frac{d\theta}{dt}=0.45 rad/hour[/tex]
