Question 68 of 82advanced🔧 ApplyNumerical4 marks

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Correct Answer

The area is decreasing at the rate of b3\sqrt{3} cm²/s.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 2 | (Page 37)
For More Understanding

Explanation

This is a related rates problem from Application of Derivatives. We need to find the rate of change of area with respect to time. The key steps involve: (1) expressing area in terms of the equal sides, (2) differentiating with respect to time using chain rule, and (3) substituting the given values. The altitude of the isosceles triangle is found using Pythagoras theorem, and then area formula is applied.

Solution Steps

  1. Step 1: Let the two equal sides of the isosceles triangle be x cm each. The base is fixed at b cm. Given: dx/dt = -3 cm/s (negative because sides are decreasing).

  2. Step 2: Find the altitude h of the triangle. Using Pythagoras theorem: h² + (b/2)² = x², so h = x2−b2/4\sqrt{x² - b²/4}.

  3. Step 3: Express the area A in terms of x. A = (1/2) × base × height = (b/2) × x2−b2/4\sqrt{x² - b²/4}.

  4. Step 4: Differentiate A with respect to time t. dA/dt = (b/2) × x × (dx/dt) / x2−b2/4\sqrt{x² - b²/4}.

  5. Step 5: Substitute x = b and dx/dt = -3. When x = b: x2−b2/4\sqrt{x² - b²/4} = b2−b2/4\sqrt{b² - b²/4} = 3b2/4\sqrt{3b²/4} = (b3\sqrt{3})/2.

  6. Step 6: Calculate dA/dt = (b/2) × b × (-3) / ((b3\sqrt{3})/2) = -b3\sqrt{3} cm²/s. The negative sign indicates area is decreasing at b3\sqrt{3} cm²/s.

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