HSC Mathematics Ext. 1 (3 Unit) SAMPLE LECTURE SLIDES HSC Exam Preparation Programs 22-26 September, 2014 c 2014 Sci SchoolTM . All rights reserved. Overview 1. Further Trigonometry 1. Further Trigonometry 2. Circle Geometry 2. Circle Geometry 3. Parametric Equations 3. Parametric Equations 4. Mathematical Induction 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 10. Rates of Change 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.1 Chain Rule Applications 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials In the 2 Unit course, we learned that ‘the rate of change’ of a function, Q(t), means ‘the derivative with respect to time’, i.e. dQ dt . In the 3 Unit course, Q may not be explicitly known as a function of t but of another variable, u, instead. Hence, dQ dt must be found using the Chain Rule. 6. Binomial Theorem dQ(u) du dQ = · dt du dt 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion • dQ Step 1: Write down the known ( du ) & unknown ( dt dt ) time-derivatives. • Step 2: Find the connection between Q and u. We need it explicitly as Q(u), with no other variables in the expression. • Step 3: Diﬀerentiate Q(u) to get dQ du . • Step 4: Substitute your values for solve for dQ dt . dQ du and du dt into the Chain Rule to © 2014 Sci School 10.1 Chain Rule Applications 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction For example, an inverted cone of base radius 6 cm and height 20 cm has water ﬂowing from its apex at the constant rate of 6 cm3 /s. Use the Chain Rule to determine the rate at which the water level is falling when the water level is 4 cm. 5. Polynomials 6 6. Binomial Theorem 7. Further Probability 8. Integration Methods r 9. Inverse Trigonometric Functions 20 h 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion • Step 1: We want dh dt when h is 4 cm. We know dV dt is −6 cm3 /s. 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.1 Chain Rule Applications 1. Further Trigonometry • Step 2: For a cone, V and h are related by V = 13 πr 2 h. 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change Since r is also a variable, we need to rearrange the expression so it contains only V and h as variables. Notice the geometric relationship between r and h in the diagram: equiangular similar triangles means the h ratio r6 is equal to 20 . Hence, h r =6× 20 Substituting this into our volume expression gives, 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion =⇒ 3h r= 10 3h 1 π 3 10 3π 3 h ∴ V = 100 V = 2 h © 2014 Sci School 10.1 Chain Rule Applications 1. Further Trigonometry • Step 3: Diﬀerentiating V = 3π 3 100 h gives us, 2. Circle Geometry dV 9π 2 = h dh 100 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions • Step 4: Substituting our values for or dh dV 100 dh = dV 9πh2 and dV dt into the Chain Rule, dh dh dV = · dt dV dt 100 dh 3 = · −6 cm /s ∴ dt 9πh2 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion Lastly, at a water level of 4 cm, our expression simpliﬁes to 200 cm3 /s dh =− 2 dt 3π (4 cm) 25 dh = − cm/s ∴ dt 6π when h = 4 cm © 2014 Sci School 10.2 Newton’s Law of Cooling 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion This is an extension on Exponential Growth and Decay discussed in the 2 Unit course. Newton showed that an object’s temperature is governed by, dT = −k(T − Tf ) dt where k > 0 is a constant and Tf is the equilibration (ﬁnal) temperature. The solution to this equation is found by integrating the inverse expression. That is, we ﬁrst invert Newton’s diﬀerential equation, dt −1 = . dT k(T − Tf ) 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.2 Newton’s Law of Cooling 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials Secondly, we integrate the fraction to form a log, −1 dT t= k(T − Tf ) −1 = ln(T − Tf ) + C. k 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion Subtracting C and multiplying by −k, give us, −kt + kC = ln(T − Tf ). We can now exponentiate both sides to arrive at e−kt+kC = T − Tf or Ae−kt = T − Tf , if we deﬁne a simpliﬁed constant as A = ekC . Solving for T gives us, T (t) = Tf + Ae−kt © 2014 Sci School 10.2 Newton’s Law of Cooling 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem The constant A represents the diﬀerence between initial and ﬁnal temperatures. To see this, substitute t = 0 into T (t) to arrive at T (0) = Tf + A, or A = T0 − Tf . T (t) T0 T (t) = Tf + Ae−kt 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change A>0 Tf 10.1 Chain Rule Applications A<0 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion T0 t 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.2 Newton’s Law of Cooling 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem For example, given a cooling constant of 0.06 min−1 , how much quicker is it to chill 18◦ C tap water to 7◦ C when using a 5◦ C refrigerator compared with a -20◦ C freezer? • Refrigerator data: T (t) = 7◦ C, T0 = 18◦ C, Tf = 5◦ C ∴ A = 13◦ C. • Step 1: Substitute the data into Newton’s temperature formula. 7 = 5 + 13e−0.06t 2 = e−0.06t ∴ 13 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion • Step 2: Solve for t by taking logs of both sides, 2 = −0.06t ln 13 2 1 ln = 31.2 min ∴ t=− 0.06 13 © 2014 Sci School 10.2 Newton’s Law of Cooling 1. Further Trigonometry • Freezer data: T (t) = 7◦ C, T0 = 18◦ C, Tf = −20◦ C ∴ A = 38◦ C. 2. Circle Geometry • Step 1: Substitute the data into Newton’s temperature formula. 3. Parametric Equations 7 = −20 + 38e−0.06t 27 = e−0.06t ∴ 38 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion • Step 2: Solve for t by taking logs of both sides, 27 = −0.06t ln 38 27 1 ln = 5.7 min ∴ t=− 0.06 38 Hence, the freezer is 26 minutes faster than the refrigerator at chilling the water from 18◦ C to 7◦ C. 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry Quesiton 1 (4 Marks) 2. Circle Geometry Alice, Bob, and Charlie stand so as to form a right-angled triangle, with Charlie and Bob forming the hypotenuse and Alice and Bob separated by 20 metres. Charlie proceeds to walk towards Alice at a constant rate of change of 2 metres per second. At what rate is the distance between Charlie and Bob changing, when Charlie is 15 metres from Alice? 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions Solution 10.1 Chain Rule Applications Let the distances between Charlie and Alice and Charlie and Bob be a and b, da respectively. We need to calculate db dt when a is 15 metres. We know that dt is −2 metres per second, since the distance is decreasing. 10.2 Newton’s Law of Cooling Using Pythagorus’ Theorem, 10. Rates of Change 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion 202 + a2 = b2 ∴ b = 400 + a2 © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion Diﬀerentiating this expression, we have 1 d db 2 2 = 400 + a da da 1 1 2 −2 = × 400 + a × 2a 2 a = √ 400 + a2 3 = , when a = 15 5 Substituting these values into the Chain Rule, we arrive at db 3 = × −2 dt 5 6 =− 5 Hence, when Charlie is 15 m from Alice, he approaches Bob at 1.2 m/s. © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry Quesiton 2 (3 Marks) 2. Circle Geometry A cup of tea has an initial temperature of 100◦ C. The temperature, T ◦ C, of the tea after t minutes is given by 3. Parametric Equations 4. Mathematical Induction T (t) = X + Y e−kt , 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions where X, Y , and k are positive constants. In a room with an ambient temperature of 19◦ C, the temperature of the tea drops to 85◦ C after 5 minutes. How long does the tea take to cool to 50◦ C? Solution The ﬁnal temperature of the tea will be 19◦ C, ∴ X = 19◦ C. Also, we are told that T (0) = 100◦ C, ∴ Y = (100 − 19)◦ C = 81◦ C. 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion When t = 5, T = 85◦ C. Substituting this into the equation, we have 80 = 19 + 81e−k×5 61 = 81e−5k 61 = e−5k 81 61 1 ∴ k = − ln 5 81 Now that all the constants are known, we can substitute T = 50◦ C. 1 5 ln( 61 81 )t 50 = 19 + 81e 1 61 31 = e 5 ln( 81 )t 81 31 ln 81 ∴ t = 1 61 = 16.9 minutes 5 ln 81 © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry Quesiton 3 (6 Marks) 2. Circle Geometry At 3pm in a school playground, a melted chocolate bar is found to be at a temperature of x◦ C. The packaging on the chocolate bar says it will begin to melt at 16◦ C. The temperature, T ◦ C, of the bar t minutes after 3pm varies according to the diﬀerential equation 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem dT 1 = ln(1.6)(A − T ), dt 60 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion where A is a constant. (i) Show that for any constant, B, a solution to the diﬀerential equation is T = A + Be 1 − 60 ln(1.6)t . ◦ (ii) After an hour, the chocolate bar increases in temperature by 15 4 C. Given the chocolate bar started to melt at 2pm, ﬁnd x and the limiting temperature of the bar. Assume the temperature of the day is constant. © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry Solution 2. Circle Geometry (i) Diﬀerentiating the proposed solution, we have 3. Parametric Equations 1 d dT = A + Be− 60 ln(1.6)t dt dt 1 1 − 60 ln(1.6)t = Be × − ln (1.6) 60 1 = (T − A) × − ln (1.6) 60 1 ln(1.6)(A − T ) = 60 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions as required. (ii) We are told that T (0) = x◦ C, ∴ x = A + B. 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry When t = −60, T = 16◦ C. Substituting these values into the equation, 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion 16 = A + Be 1 − 60 ln(1.6)×−60 16 = A + Beln(1.6) 16 = A + B × 1.6 16 − A ∴ B= 1.6 When t = 60, T = (x + 15 ◦ 4 ) C. Hence, we have 15 4 15 A+B+ 4 15 B+ 4 ∴ B x+ 1 = A + Be− 60 ln(1.6)×60 = A + Be =B× = −10 − ln( 16 10 ) 10 16 © 2014 Sci School 10.3 HSC-Adapted Questions 1. Further Trigonometry Combining these equations, we have 16 − A = −10 1.6 ∴ A = 16 + 16 = 32 2. Circle Geometry 3. Parametric Equations 4. Mathematical Induction 5. Polynomials 6. Binomial Theorem 7. Further Probability 8. Integration Methods 9. Inverse Trigonometric Functions 10. Rates of Change 10.1 Chain Rule Applications 10.2 Newton’s Law of Cooling 10.3 HSC-Adapted Questions 11. Rectilinear Motion 12. Projectile Motion 13. Simple Harmonic Motion Since x = A + B, we have x = 32 − 10 ∴ x = 22 1 The limiting temperature occurs when t → ∞. Since e− 60 ln(1.6)t → 0 as t → ∞, we arrive at T (t → ∞) = 32 − 10 × 0 = 32 which is the equilibrium temperature. © 2014 Sci School

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