Question

The base of the right triangle in the figure increases at a rate of 4 cm/s, while the height remains constant at ℎ=20 cm. How fast is the angle changing when =32?

Answer 1

The rate at which the angle changes when the base is equal to 32 is:

a' = 2.876 /s.

How fast is the angle changing?

We have a right triangle where the two catheti measure:

• b = 32cm + 4cm/s*t
• h = 20cm

(We don't know which angle we want, I assume that we want the angle adjacent to the base). So we can use the trigonometric relation:

tan(a) = (opposite cathetus)/(adjacent cathetus).

In this case, we will have:

• Opposite cathetus = 20cm
• Adjacent cathetus = 32cm + 4cm/s*t

tan(a) = 20cm/(32cm + 4cm/s*t)

a = Atan(20cm/(32cm + 4cm/s*t))

Now we need to differentiate it with respect to x, remember that:

$\frac{dAtan(x)}{dx} = \frac{1}{1 + x^2}$

Then we will have:

$a' = \frac{4cm/s}{1 + (20cm/(32cm + 4cm/s*t)^2}$

The rate of change when b = 32, is what we get when we have t = 0s, replacing that we get:

$a' = \frac{4cm/s}{1 + (20cm/(32cm))^2} = 2.876 /s$

If you want to learn more about rates of change, you can read:

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