Respuesta :
Answer:
[tex]\frac{d\theta}{dt}=-149.7735[/tex] negative sign indicates decreasing.
Explanation:
Given:
- [tex]x=9\ units[/tex]
- [tex]y=5\ units[/tex]
- [tex]\frac{dx}{dt} =9\ units.s^{-1}[/tex]
- [tex]\frac{dy}{dt} =-19\ units.s^{-1}[/tex]
Using trigonometric relation of the sides and angles of a right angled triangle:
[tex]tan\ \theta=\frac{y}{x}[/tex] ...........................(1)
Differentiate this eq. w.r.to time 't':
[tex]\frac{d}{dt} (tan\ \theta)=\frac{d}{dt}( \frac{y}{x})[/tex]
[tex]sec^2\ \theta\ \frac{d\theta}{dt} =\frac{x\frac{dy}{dt}-y\frac{dx}{dt} }{x^2}[/tex]
[tex]\frac{d\theta}{dt}=\frac{cos^2\ \theta}{x^2} (x\frac{dy}{dt}-y\frac{dx}{dt})[/tex] ............(2)
Now , putting the respective values in eq. (2)
[tex]\frac{d\theta}{dt}=\frac{cos^2\ \theta}{9^2} (9\times(-19)-5\times5)[/tex]
[tex]\frac{d\theta}{dt}=\frac{-196\ cos^2\ \theta}{81}[/tex] ......................(3)
Now, determine cos θ from the triangle:
[tex]cos\ \theta =\frac{x^2}{\sqrt{x^2+y^2} }[/tex]
[tex]cos\ \theta =\frac{9^2}{\sqrt{9^2+5^2} }[/tex]
[tex]cos\ \theta =\frac{81}{\sqrt{81+25} }[/tex]
[tex]cos\ \theta =7.8674[/tex]
Putting this value in eq. (3)
[tex]\frac{d\theta}{dt}=\frac{-196\times 61.896}{81}[/tex]
[tex]\frac{d\theta}{dt}=-149.7735[/tex] negative sign indicates decreasing.
Answer:
Explanation:
According to question
tan θ = y / x
Differentiate with respect to t on both the sides
[tex]Sec^{2}\theta \times \frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}}[/tex]
[tex]\frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}\times Sec^{2}\theta}[/tex] .... (1)
According to question,
tan θ = 5 / 9
So, Sec θ = 10.3 / 9 = 1.14
dx/dt = 9 units/s
dy/dt = 19 units/s
Substitute the values in equation (1), we get
[tex]\frac{d\theta }{dt}=\frac{9\times 19-5\times 9}{81\times 1.14^{2}}[/tex]
dθ/dt = 1.2 units/s
