Case Based Questions - Worksheet 1

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Chapter 5 Class 12 - Continuity & Differentiability - Case Based Question Worksheet 1 by teachoo Chapter: Chapter 5 Class 12 - Continuity & Differentiability Name: _____________________________ School: _____________________________ Roll Number: _____________________________ Case Study 1: Rocket Launch The altitude (height) of a rocket, h (in meters), t seconds after launch is given by the piecewise function: h(t)={■(2t^2&" if " 0≤t≤20@100t-1200&" if " t>20)┤ Continuity Check: Does the rocket's altitude change continuously at t=20 seconds? Show your work. Velocity: The velocity v(t) is the derivative of the altitude, h^' (t). Find the velocity of the rocket for 0<t<20 and for t>20. Differentiability: Is the velocity function continuous at t=20 ? In other words, is the altitude function h(t) differentiable at t=20 ? What does this physically mean for the rocket? Acceleration: The acceleration a(t) is the derivative of velocity, v^' (t). What is the rocket's acceleration before and after 20 seconds? What event could this change in acceleration represent? Case Study 2: Smart Thermostat A smart thermostat maintains the temperature in a room. The temperature T (in ^∘ C ) is given by T(t)=21-3cos⁡(πt/12), where t is the time in hours after midnight. Rate of Change: Find a function for the rate at which the temperature is changing with respect to time, T^' (t). Maximum Rate of Cooling: At what time is the room cooling down the fastest (i.e., when is T^' (t) at its most negative value)? Constant Temperature: At what times is the rate of change of temperature equal to zero? What does this signify about the temperature at these times? Second Derivative: Find the second derivative, T^'' (t). What does the sign of T^'' (t) tell you about whether the rate of temperature change is increasing or decreasing? Case Study 3: Manufacturing Cost The cost C (in thousands of dollars) to produce x hundred units of a product is given by the function C(x)=√(x^2+900). The company currently produces 40 hundred (4000) units. Current Cost: What is the current cost to produce 4000 units? Marginal Cost: The "marginal cost" is the instantaneous rate of change of cost, C^' (x). Find the marginal cost function. Application of Derivative: What is the marginal cost when the company is producing 4000 units? Interpret this value in the context of the business. Implicit Relation: If the revenue is related to the cost by the implicit equation R^2-C^2=1600, find the rate of change of revenue with respect to the number of units, dR/dx, when the cost is the value you found in question 1. Important links Answer of this worksheet - https://www.teachoo.com/25578/5355/Case-Based-Questions---Worksheet-1/category/Teachoo-Questions---Case-Based/ Full Chapter with Explanation, Activity, Worksheets and more – https://www.teachoo.com/subjects/cbse-maths/class-12th/ Maths Class 12 - https://www.teachoo.com/subjects/cbse-maths/class-12th/ For more worksheets, ad-free videos and Sample Papers – subscribe to Teachoo Black here - https://www.teachoo.com/black/   Answer Key to Case Based Question Worksheet 1 Case Study 1: Rocket Launch Continuity: h(20)=2(20)^2=800⋅lim_(t→20^+ ) h(t)=100(20)-1200=800. Since h(20)=lim_(t→20^+ ) h(t), the function is continuous. Velocity: For 0<t<20,v(t)=h^' (t)=4t. For t>20,v(t)=h^' (t)=100. Differentiability: The left-hand derivative at t=20 is 4(20)=80. The right-hand derivative is 100 . Since 80≠100, the function is not differentiable. This means the rocket experiences a sudden change in velocity (a "jerk"), implying an instantaneous change in acceleration. Acceleration: For 0<t<20,a(t)=v^' (t)=4" " m/s^2. For t>20,a(t)=v^' (t)=0" " m/s^2. This could represent the first-stage engine cutting off. Case Study 2: Smart Thermostat Rate of Change: T^' (t)=-3(-sin⁡(πt/12))⋅π/12=π/4 sin⁡(πt/12). Maximum Rate of Cooling: Cooling is fastest when T^' (t) is most negative. This occurs when sin⁡(πt/12)=-1, which happens at πt/12=3π/2, so t=18 (6 PM). Constant Temperature: T^' (t)=0 when sin⁡(πt/12)=0. This occurs at t=0,12,24. This signifies the temperature is at its minimum (at t=0,24 ) or maximum (at t=12 ) and is momentarily not changing. Second Derivative: T^'' (t)=π/4 cos⁡(πt/12)⋅π/12=π^2/48 cos⁡(πt/12). When T^''>0, the rate of change is increasing. When T^''<0, the rate of change is decreasing. Case Study 3: Manufacturing Cost Current Cost: C(40)=√(40^2+900)=√(1600+900)=√2500=50. The cost is $50,000. Marginal Cost: C^' (x)=1/(2√(x^2+900))⋅2x=x/√(x^2+900). Application: C^' (40)=40/√(40^2+900)=40/50=0.8. This means at a production level of 4000 units, the cost to produce the next 100 units is approximately 0.8 thousand dollars, or $800. Implicit Relation: Differentiating R^2-C^2=1600 w.r.t. x gives 2R dR/dx-2C dC/dx=0, so dR/dx=C/R dC/dx. When C=50,R^2=1600+ 50^2=4100, so R=√4100. Thus, dR/dx=50/√4100(0.8)≈0.624.