Related Rates Math Example 2

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Example 2

hard
A 13-ft ladder rests against a wall. The base slides away at 5 ft/s. How fast is the top sliding down when the base is 5 ft from the wall?

Solution

  1. 1
    Let xx = distance of base from wall, yy = height of top. Pythagorean: x2+y2=169x^2 + y^2 = 169.
  2. 2
    When x=5x = 5: y=169โˆ’25=12y = \sqrt{169 - 25} = 12 ft.
  3. 3
    Differentiate with respect to tt: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.
  4. 4
    Substitute: 2(5)(5)+2(12)dydt=0โ‡’50+24dydt=02(5)(5) + 2(12)\frac{dy}{dt} = 0 \Rightarrow 50 + 24\frac{dy}{dt} = 0.
  5. 5
    dydt=โˆ’5024=โˆ’2512\frac{dy}{dt} = -\frac{50}{24} = -\frac{25}{12} ft/s.

Answer

The top slides down at 2512โ‰ˆ2.08\dfrac{25}{12} \approx 2.08 ft/s.
Never substitute numerical values before differentiating. The Pythagorean theorem links xx and yy; differentiating links dx/dtdx/dt and dy/dtdy/dt. The negative sign confirms the top is moving downward.

About Related Rates

Problems where two or more quantities change with time and are related by an equation. Differentiate the equation with respect to time tt and use known rates to find an unknown rate.

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