Question

Help me with differential calculus. I am too dumb :(

The figure above represents an observer at point A watching balloon B as it rises from point C. The balloon is rising at a constant rate of 3 meters per second and the observer is 100 meters from point C. Find the rate of change in θ at the instant when y = 50

Answer 1

I did\[\tan \theta = \frac{ y }{ 100 }\]I then took the derivative in terms of time\[\sec^2 \theta \frac{ d \theta }{ dt } = \frac{ 1 }{ 100 }\frac{ dy }{ dt }\] Using the law of sines, I figured out that the angle \[\theta = 26.865 °\] I plugged in the values to get \[\sec^2(26.865)\frac{ d \theta }{ dt } = \frac{ 1 }{ 100 }(3)\] I am wondering if I am doing this right so far...

Answer 2

@inkyvoyd @IrishBoy123 @Directrix

Answer 3

\[\sec ^2 \theta \frac{ d \theta }{ dt }=\frac{ 1 }{ 100 }\frac{ dy }{ dt }\]
\[\left( 1+\tan ^2 \theta \right)\frac{ d \theta }{ dt }=\frac{ 1 }{ 100 }\frac{ dy }{ dt }\]
\[\left( 1+\frac{ y^2 }{ 100^2 } \right)\frac{ d \theta }{ dt }=\frac{ 1 }{ 100 }\frac{ dy }{ dt }\]
\[\frac{ dy }{ dt }=3 m/s\]
when y=50
\[\frac{ d \theta }{ dt }=?\]

Answer 4

Why was it necessary to change the sec^2 to 1 + tan^2 ?

Answer 5

no need to find value of theta in degrees.

Answer 6

I got 0.024 radians/s
Is this right?

Answer 7

welcome to os