Question

Area Let \(\theta\) be the angle between equal sides of an isosceles triangle and let \(x\) be the length of these sides. \(x\) is increasing at \(\frac{1}{2}\) meter per hour and \(\theta\) is increasing at \(\pi / 90\) radian per hour. Find the rate of increase of the area when \(x=6\) and \(\theta=\pi / 4\).

Short answer

The rate of increase of the area of the isosceles triangle when \(x=6\) and \(\theta=\pi / 4\) is \(\frac{3}{\sqrt{2}}+3\sqrt{2}\pi\).

Step-by-step solution

Derive the formula for the area of an isosceles triangle

Remember, the formula for the area (A) of an isosceles triangle is given by \(A=\frac{1}{2}x^2 \sin(\theta)\)

Differentiate the formula for the area with respect to time (t)

Here, it is necessary to apply the chain rule. Differentiate \(A\) with respect to \(x\) and \(\theta\), then multiply by the rate of change of \(x\) and \(\theta\) respectively. The derivative will be \(\frac{dA}{dt} = x\sin(\theta)\frac{dx}{dt}+\frac{1}{2}x^2 \cos(\theta)\frac{d\theta}{dt}\)

Substitute the given values

Now substitute \(x=6\), \(\sin(\theta)= \sin(\frac{\pi}{4})= \frac{1}{\sqrt{2}}\), \(\cos(\theta) = \cos(\frac{\pi}{4})= \frac{1}{\sqrt{2}}\), \(\frac{dx}{dt} = \frac{1}{2}\), and \(\frac{d\theta}{dt}= \frac{\pi}{90}\) into the derivative.

Compute the result

After substituting these values, compute the rate of change of \(A\). The result will be \(\frac{dA}{dt} = 6\cdot \frac{1}{\sqrt{2}}\cdot \frac{1}{2} + \frac{1}{2}\cdot 6^2 \cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{90}\). Simplify this to get the final result.