related rates?

Question

a balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. find the rate of change of the angle of elevation of hte balloon from the observer when thte balloon is 30 meters above ground.

thank you.

Answer 1

When the balloon is 30 meters above the ground, you can use a right triangle to find the distance from the balloon and the observer. Since the height is 30 meters and the distance from the observer to the balloon on ground is also 30 meters, you have a right triangle with both legs = 30 meters.

You can now deduce an equation to relate the sides of the triangle with the angle of elevation. Using simple trigonometry, you have:

tan(x) = (height)/(ground distance)

Taking the derivative of both sides with respect to 't', you get:

sec^2(x)(dx/dt) = (1/ground distance)*(dh/dt), and we know that sec(x) = 1/cos(x). Therefore, this can be simplified to:

dx/dt = [cos^2(x)]/(ground distance)*(dh/dt). Since we do not know the angle 'x', we can re-write the cosine relationship using trig. We know that:

cos(x) = (ground distance)/(hypotenuse). The hypotenuse is found using the 45:45:90 special right triangle. In this relationship, the sides of the triangle have the relationship x:x:x*sqrt(2). Therefore, we can conclude that the hypotenuse is 30*sqrt(2) = 42.43 m

Now you can simplify:

dx/dt = (ground distance/hypotenuse)^2/(ground distance)*(dh/dt)

= (ground distance)/(hypotenuse)^2*(dh/dt). Plug in our known values and solve for dx/dt

dx/dt = (30)*(3 m/s)/[(30*sqrt(2))^2]

dx/dt = 0.05 degree/second

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Hope this helps

Answer 2

Right triangle. Angle = arctangent of height divided by horizontal range.

Take the derivative. Plug in the rate and other parameters.