The setup steps and common scenarios for related rates and optimization word problems in calculus. Front: the scenario type or step. Back: how to set up or solve it.
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- First step when solving a related rates problem
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Draw a diagram and label all variables, including time as the independent variable. Identify which rate you need to find and which rates are given- Why differentiate both sides of an equation with respect to time in related rates
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Because the rates of change of the variables are related through the equation. Differentiating with respect to time (d/dt) captures how all variables change together- Key principle when differentiating an equation in related rates
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Use implicit differentiation with respect to time. Treat all variables except time as functions of time, applying the chain rule to every term- When to substitute known values in a related rates problem
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After differentiating with respect to time but before solving for the unknown rate. Substituting first can eliminate variables you still need- Ladder sliding down a wall (base moves 2 ft/s): how to set up
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Let x = distance from wall to base, y = height on wall. Given dx/dt = 2. Use x^2 + y^2 = L^2 (L is ladder length). Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. Solve for dy/dt- Water pouring into a cone at constant rate: how to find water level rise
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Let h = height, r = radius at surface. Use cone geometry r/h = R/H (cone ratio). Write V = (1/3)pi*r^2*h in terms of h only. Differentiate dV/dt and solve for dh/dt- Shadow cast by a person walking away from a lamp: how to set up
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Let h = lamp height, p = person height, x = distance from lamp to person, s = distance from lamp to shadow tip. Use similar triangles: h/s = p/(s-x). Differentiate with respect to time to relate dx/dt to ds/dt- Circular ripple expanding on water: radius grows at constant rate, find area increase
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Let r = radius, A = area, given dr/dt = constant. Use A = pi*r^2. Differentiate: dA/dt = 2*pi*r*(dr/dt). Substitute known values to find dA/dt- Two cars approaching intersection from perpendicular directions: find distance change rate
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Let x = distance of car 1 from intersection, y = distance of car 2, d = distance between them. Use d^2 = x^2 + y^2. Differentiate: 2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt). Solve for dd/dt- Spherical balloon expands: radius increases 3 cm/s, how fast does volume increase
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Let r = radius, V = volume, dr/dt = 3. Use V = (4/3)*pi*r^3. Differentiate: dV/dt = 4*pi*r^2*(dr/dt). Substitute r and dr/dt to find dV/dt- First step in an optimization problem
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Identify what quantity you want to optimize (maximize or minimize) and write it as a function of one or more variables. This is your objective function- What is a constraint in an optimization problem
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A limitation or relationship that restricts which values the variables can take. Constraints often allow you to write the objective function in terms of one variable instead of multiple- How to find candidate points for optimization
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Find critical points by setting the first derivative equal to zero and solving. Also evaluate the objective function at any boundary points (endpoints of the domain)- First derivative test in optimization
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Check the sign of the derivative before and after each critical point. If it changes from positive to negative, the point is a local maximum. If it changes from negative to positive, it is a local minimum- Second derivative test at a critical point
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At critical point c where f'(c) = 0: if f''(c) > 0, c is a local minimum; if f''(c) < 0, c is a local maximum; if f''(c) = 0, the test is inconclusive
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