Differential Calculus II: Applications in Curve Analysis
This content discusses various applications of differential calculus in analyzing curves, including finding slopes, turning points, and maximum heights. Detailed solutions and explanations are provided for revision and exam-style questions related to curve equations. The content covers topics such as determining increasing or decreasing behavior, calculating tangent slopes, locating turning points, and sketching curves based on equations provided.
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CHAPTER 27 Differential Calculus II Applications Solutions: Revision - Section A
27 Revision and Exam Style Questions: Section A 1. Curve has equation y = 1 + x 2x2. Find the slope of the tangent to the curve at the points ( 2, 9) and (2, 5). Hence, decide if the graph is increasing or decreasing at each point. (i) y = 1 + x 2x2 dy dx= Slope of tangent = 1 4 x x= 2: Slope = 1 4( 2) = 1 + 8 = 9 This is a positive value, therefore the graph is increasing at this point.
27 Revision and Exam Style Questions: Section A 1. Curve has equation y = 1 + x 2x2. Find the slope of the tangent to the curve at the points ( 2, 9) and (2, 5). Hence, decide if the graph is increasing or decreasing at each point. (i) dy dx= Slope of tangent = 1 4 x x = 2: Slope = 1 4(2) = 1 8 = 7 This is a negative value, therefore the graph is decreasing at this point
27 Revision and Exam Style Questions: Section A 1. Curve has equation y = 1 + x 2x2. Find the coordinates of the local turning point on the curve. (ii) dy dx=0 At the turning points Find y coordinate: y =1+x-2x2 y = 1+x-2x2 2 -21 =1+1 dy dx=1-4x 4 4 =1+1 1 1 4x = 0 4-2 16 -4x=-1 =1+1 4-1 x =1 8 4 =9 Coordinates of the turning point: 1 4, 9 8 8
27 Revision and Exam Style Questions: Section A 1. Curve has equation y = 1 + x 2x2. Hence, draw a rough sketch of the curve. (iii) y =1+x-2x2 2 shaped curve since the term is negative x Points on curve (-2,-9) and (2,-5) Max point 1 4,11 8 ?
27 Revision and Exam Style Questions: Section A 2. The height, hmetres, of a balloon is related to the time t seconds by h = 90t 10t2. Find: its height after 2 seconds (i) h = 90t 10t2 t = 2: h= 90(2) 10(2)2 = 180 10(4) = 180 40 = 140 metres
27 Revision and Exam Style Questions: Section A 2. The height, hmetres, of a balloon is related to the time t seconds by h = 90t 10t2. Find: the maximum height reached by the balloon. (ii) At maximum height, dh dt=0: h= 90t-10t2 dh dt=90-20t 0=90-20t 20t =90 t =4 5 second
27 Revision and Exam Style Questions: Section A 2. The height, hmetres, of a balloon is related to the time t seconds by h = 90t 10t2. Find: the maximum height reached by the balloon. (ii) Find h at this time: t= 4 5: h = 90t 10t2 = 90(4 5) 10(4 5)2 = 405 202 5 h = 202 5 metres (Maximum height)
27 Revision and Exam Style Questions: Section A y =-2x2+7x-5, find dy dx. 3. (i) If y= 2x2 + 7x 5 dy dx= + 4 7 x
27 Revision and Exam Style Questions: Section A 3. (ii) Find the equation of the tangent to the curve at the point where x= 0 Slope at x = 0:dy dx=-4(0)+7 =0+7 Slope=7 y value when x = 0: y= 2x2 + 7x 5 y= 2(0)2+ 7(0) 5 y= 0 + 0 5 y= 5
27 Revision and Exam Style Questions: Section A 3. (ii) Find the equation of the tangent to the curve at the point where x= 0 Point on tangent (0, 5) (0, 5) x1, y1 m = 7 Equation of tangent: y y1 = m(x x1) y + 5 = 7(x 0) y + 5 = 7x 7x y 5 = 0
27 Revision and Exam Style Questions: Section A 3. (iii) Find the equation of the tangent to the curve at the point where x= 3. dy dx= + Slope at x = 3: 4 7 x = 4(3) + 7 = 12 + 7 Slope = 5
27 Revision and Exam Style Questions: Section A 3. (iii) Find the equation of the tangent to the curve at the point where x= 3. y value when x = 3: (3, 2) x1, y1 m= 5 y= 2x2 + 7x 5 y y1 = m(x x1) Equation: y= 2(3)2+ 7(3) 5 y+ 2 = 5(x 3) y= 18 + 21 5 y+ 2 = 5x + 15 y= 2 5x + y+ 2 15 = 0 (3, 2) 5x + y 13 = 0
27 Revision and Exam Style Questions: Section A 4. Once an aeroplane touches down, it applies its brakes. The distance, sin metres, which it has travelled along the runway at time t seconds after touchdown is given by s(t) = 200t 4t2. Find: the speed of the aeroplane at t = 3 seconds (i) s(t) = 200t 4t2 Speed = s (t) = 200 8t s (3) = 200 8(3) t = 3: = 200 24 = 176 m/sec
27 Revision and Exam Style Questions: Section A 4. Once an aeroplane touches down, it applies its brakes. The distance, sin metres, which it has travelled along the runway at time t seconds after touchdown is given by s(t) = 200t 4t2. Find: the constant deceleration of the aeroplane (ii) Speed = s (t) = 200 8t Deceleration = s (t) = 8 m/sec2
27 Revision and Exam Style Questions: Section A 4. Once an aeroplane touches down, it applies its brakes. The distance, sin metres, which it has travelled along the runway at time t seconds after touchdown is given by s(t) = 200t 4t2. Find: the time taken for the aeroplane to come to rest (iii) Come to rest is when speed = 0 Speed = s (t) = 200 8t = 0 8t= 200 8t = 200 t =200 8 t = 25 sec
27 Revision and Exam Style Questions: Section A 4. Once an aeroplane touches down, it applies its brakes. The distance, sin metres, which it has travelled along the runway at time t seconds after touchdown is given by s(t) = 200t 4t2. Find: the distance travelled by the aeroplane before coming to rest. (iv) Find the distance travelled when t = 25 sec Distance = s(t) = 200t 4t2 = 200(25) 4(25)2 = 5000 2500 = 2500 metres
27 Revision and Exam Style Questions: Section A 5. For what values of x do the tangents to the graph of y = x2 + 7x 11 have negative slopes? y = x2 + 7x 11 dy dx= 2x+7 dy dx<0 For negative slopes: 2x+7<0 2x<-7 x <-7 2 x<-3 5
27 Revision and Exam Style Questions: Section A 6. Find the equation of the tangent to the curve f(x) = 2x2 3x + 5, at the point ( 1, 10). f (x) = 2x2 3x + 5 f (x) = 4x 3 x= 1:f ( 1) = 4( 1) 3 slope = 4( 1) 3 slope = 4 3 slope = 7 ?
27 Revision and Exam Style Questions: Section A 6. Find the equation of the tangent to the curve f(x) = 2x2 3x + 5, at the point ( 1, 10). ( 1, 10) x1, y1 m= 7 y y1 = m(x x1) Equation: y 10 = 7(x + 1) y 10 = 7x 7 7x + y 10 + 7 = 0 7x + y 3 = 0 ?
27 Revision and Exam Style Questions: Section A 7. Find the point on the curve y = x2 + 12x + 6 at which the slope of a tangent is 2. y = x2 + 12x + 6 Find y-value when x= 5 dy dx y = x2 + 12x + 6 slope of tangent = = 2x + 12 = ( 5)2+ 12( 5) + 6 2x + 12 = 2 slope = 2 : = 25 60 + 6 2x= 2 12 = 29 2x= 10 Point ( 5, 29) x= 5
27 Revision and Exam Style Questions: Section A 8. A train is travelling along a track. Suddenly, the brakes are applied. From the time the brakes are applied (t= 0 seconds), the distance travelled by the train, in metres, is given by s=22t-1 5t2 What is the speed of the train at the moment the brakes are applied? (i) s=22t-1 5t2 dt=22-2 Speed =ds 5t
27 Revision and Exam Style Questions: Section A 8. A train is travelling along a track. Suddenly, the brakes are applied. From the time the brakes are applied (t= 0 seconds), the distance travelled by the train, in metres, is given by s=22t-1 5t2 What is the speed of the train at the moment the brakes are applied? (i) Brakes are applied when t = 0: Speed =22-2 5(0) = 22 0 = 22 m/sec
27 Revision and Exam Style Questions: Section A 8. A train is travelling along a track. Suddenly, the brakes are applied. From the time the brakes are applied (t= 0 seconds), the distance travelled by the train, in metres, is given by s=22t-1 5t2 How many seconds does it take for the train to come to rest? (ii) Train comes to rest when speed = 0 = 22-2 Speed =ds 5t dt 2 5t = 22 0 ( 5) 110 2t = 0 2t= 110 t = 55 seconds
27 Revision and Exam Style Questions: Section A 8. A train is travelling along a track. Suddenly, the brakes are applied. From the time the brakes are applied (t= 0 seconds), the distance travelled by the train, in metres, is given by s=22t-1 5t2 How far does the train travel in that time? (iii) Find s when t = 55: s=22t-1 5t2 s=22(55)-1 5(55)2 s=1210-605 s=605 metres
27 Revision and Exam Style Questions: Section A 8. A train is travelling along a track. Suddenly, the brakes are applied. From the time the brakes are applied (t= 0 seconds), the distance travelled by the train, in metres, is given by s=22t-1 5t2 Find the deceleration the train undergoes. (iv) = 22-2 Speed =ds 5t dt Deceleration =d2s dt2=-2 5 m/s2
27 Revision and Exam Style Questions: Section A 9. A farmer has 80 m of fencing with which to make a rectangular field along the side of a river. The river will form one side of the field. The width of the two shortest sides is W. Find the length of the field, in terms of W. (i) Perimeter = L + W + W = 80 L + 2W = 80 L= 80 2W ?
27 Revision and Exam Style Questions: Section A 9. A farmer has 80 m of fencing with which to make a rectangular field along the side of a river. The river will form one side of the field. The width of the two shortest sides is W. Show that the area of the field, A, is given by A = 80W 2W2 (ii) Area = L W Area = (80 2W)(W) Area = 80W 2W2 ?
27 Revision and Exam Style Questions: Section A 9. A farmer has 80 m of fencing with which to make a rectangular field along the side of a river. The river will form one side of the field. The width of the two shortest sides is W. Hence, find the maximum area that can be enclosed. (iii) dA dW=0: Maximum area when A = 80W 2W2 dA dW=80-4W 80 4W = 0 (at max) 4W= 80 W = 20 m ?
27 Revision and Exam Style Questions: Section A 9. A farmer has 80 m of fencing with which to make a rectangular field along the side of a river. The river will form one side of the field. The width of the two shortest sides is W. Hence, find the maximum area that can be enclosed. (iii) Find area when W = 20 m: Maximum Area = 80W 2W 2 = 80(20) 2(20)2 = 1600 2(400) = 1600 800 = 800 m2 ?
