Math Problem Statement

The base of a right triangle is fixed at 4 meters. The height y is increasing at a rate of 2m/s when y = 3 meters. Determine the rate at which theta is increasing, in radians per second, at this time.

Solution

To find the rate at which the angle θ\theta is increasing in a right triangle with a fixed base and a variable height, we can use trigonometry and related rates.

Step-by-Step Solution:

  1. Identify Variables and Known Rates:

    • Base of the triangle (xx) is fixed at 44 meters.
    • Height (yy) is variable and is increasing at dydt=2m/s\frac{dy}{dt} = 2 \, \text{m/s}.
    • At the instant we are considering, y=3y = 3 meters.
  2. Set Up the Trigonometric Relation:

    • In a right triangle, we can express tan(θ)\tan(\theta) in terms of yy and xx: tan(θ)=yx\tan(\theta) = \frac{y}{x}
    • Since xx is fixed, this relationship simplifies to θ\theta being a function of yy alone.
  3. Differentiate with Respect to Time:

    • We differentiate tan(θ)=yx\tan(\theta) = \frac{y}{x} with respect to time tt: sec2(θ)dθdt=1xdydt\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{x} \cdot \frac{dy}{dt}
    • Solving for dθdt\frac{d\theta}{dt}: dθdt=1xsec2(θ)dydt\frac{d\theta}{dt} = \frac{1}{x \cdot \sec^2(\theta)} \cdot \frac{dy}{dt}
  4. Calculate sec2(θ)\sec^2(\theta):

    • Since tan(θ)=yx=34\tan(\theta) = \frac{y}{x} = \frac{3}{4} at the given time, we can find θ\theta using the tangent inverse: θ=tan1(34)\theta = \tan^{-1}\left(\frac{3}{4}\right)
    • Now, sec2(θ)=1+tan2(θ)\sec^2(\theta) = 1 + \tan^2(\theta): sec2(θ)=1+(34)2=1+916=2516\sec^2(\theta) = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}
  5. Substitute Values:

    • Substitute x=4x = 4, sec2(θ)=2516\sec^2(\theta) = \frac{25}{16}, and dydt=2\frac{dy}{dt} = 2 into the differentiated equation: dθdt=1425162\frac{d\theta}{dt} = \frac{1}{4 \cdot \frac{25}{16}} \cdot 2 Simplifying: dθdt=1100162=161002=32100=0.32radians per second\frac{d\theta}{dt} = \frac{1}{\frac{100}{16}} \cdot 2 = \frac{16}{100} \cdot 2 = \frac{32}{100} = 0.32 \, \text{radians per second}

Answer:

The rate at which θ\theta is increasing at this time is dθdt=0.32radians per second\frac{d\theta}{dt} = 0.32 \, \text{radians per second}.

Would you like a breakdown of any specific step, or do you have additional questions?

Here are some related questions for further exploration:

  1. How would the answer change if the base length was variable instead of fixed?
  2. What if the rate of height increase was a function of time?
  3. Can this problem be solved using a different trigonometric function?
  4. How does the result change if the triangle’s base is extended to a larger fixed length?
  5. What implications would this rate of angle change have in real-world applications like tilting structures?

Tip: When dealing with related rates, always differentiate with respect to time, even if the initial relationship doesn’t directly involve time.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Related Rates
Calculus

Formulas

tan(θ) = y / x
sec^2(θ) = 1 + tan^2(θ)
dθ/dt = (1 / (x * sec^2(θ))) * (dy/dt)

Theorems

Trigonometric identity
Related rates theorem

Suitable Grade Level

Grades 11-12