# 2) A hot air balloon rising vertically is being tracked by an observer on the ground, located 60 meters from the take-off point. At a certain instant, the balloon is 30 meters above the ground and the angle of elevation (between the ground and the observer’s sight line) is increasing at the rate of 0. 1 rad./sec. How fast is the balloon rising at that instant?

## let the height of the balloon be h m

then tanØ = h/60 , where Ø is the angle of elevation

h = 60 tanØ

dh/dt = sec^2 Ø dØ/dt

when h = 30, dØ/dt = .1 rad/sec

then tanØ = 30/60 = 1/2

hypotenuse = 30√5

secØ = 30√5/60 = √5/2

sec^2 Ø = 5/4

dh/dt = (5/4)(.1) m/s

= .125 m/sec or 12.5 cm/second

check my arithmetic

## To find how fast the balloon is rising at a certain instant, we can use trigonometry and calculus.

Let's denote the height of the balloon above the ground as h (in meters) and the distance from the observer to the take-off point as x (in meters).

From the given information, we know that x = 60m and h = 30m. We want to find dh/dt, the rate at which the balloon is rising (in meters per second), when the angle of elevation is increasing at a rate of 0.1 rad/sec.

To determine the relationship between x, h, and the angle of elevation, we can use the tangent function. Tangent is the ratio of the opposite side to the adjacent side in a right triangle. In this case, the opposite side is h, and the adjacent side is x. Therefore, we have:

tan(theta) = h / x

Rearranging the formula, we get:

h = x * tan(theta) ------ (Equation 1)

Now, let's differentiate Equation 1 with respect to time (t):

dh/dt = dx/dt * tan(theta) + x * d(tan(theta))/dt

The first term on the right side of the equation represents the rate of change of x with respect to time, which is 0 since the observer is stationary. The second term represents the rate of change of tangent with respect to time, which is given as 0.1 rad/sec.

dh/dt = x * (0.1) [tan(theta) * sec^2(theta)] ------ (Equation 2)

To calculate dh/dt, substitute the known values of x = 60m, theta = arctan(h/x), and d(tan(theta))/dt = 0.1 rad/sec into Equation 2. Then solve for dh/dt:

theta = arctan(h/x)

h/x = tan(theta)

h/60 = tan(theta)

30/60 = tan(theta)

0.5 = tan(theta)

theta = arctan(0.5)

Now, calculate d(tan(theta))/dt at theta = arctan(0.5) using the derivative of the tangent function:

d(tan(theta))/dt = sec^2(theta) * d(theta)/dt

= sec^2(arctan(0.5)) * 0.1 rad/sec

Using trigonometric identities, we know that sec^2(arctan(0.5)) = 1 / (cos^2(arctan(0.5))).

To find cos(arctan(0.5)), we can use the Pythagorean identity: sin^2(arctan(x)) + cos^2(arctan(x)) = 1.

In this case, sin(arctan(0.5)) = 0.5 since sin is the ratio of the opposite side to the hypotenuse, and in this case, the opposite side is 0.5 and the hypotenuse is 1.

Using the Pythagorean identity, we have:

sin^2(arctan(0.5)) + cos^2(arctan(0.5)) = 1

(0.5)^2 + cos^2(arctan(0.5)) = 1

0.25 + cos^2(arctan(0.5)) = 1

cos^2(arctan(0.5)) = 0.75

cos(arctan(0.5)) = sqrt(0.75)

Now, substitute the values back into Equation 2:

dh/dt = (60) * (0.1) * [tan(arctan(0.5)) * (1 / sqrt(0.75))]

= 6 * [0.5 * (1 / sqrt(0.75))]

= 6 * 0.5 * (sqrt(4/3)/sqrt(1))

= 6 * 0.5 * (2/sqrt(3))

= 6/sqrt(3)

= 2 * sqrt(3) m/s

Therefore, at that instant, the balloon is rising at a rate of 2 * sqrt(3) m/s.