WorkSHEET 2.1 Applications of differentiation Name:

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Franklin Flowers
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1 WorkSHEET. Applications o dierentiation Name: The displacement o a vehicle can be described by the unction; (d in metres and t in seconds) d t = 4t + How ast is the car travelling? The derivative o a displacement time graph gives speed. d ' t = 4 thereore the vehicle is travelling at 4 metres per second. A rocket blasts o and its height (metres) ater t (seconds) is modeled by the unction; and clearly; h t = t 0 t a) What is its initial height? b) What is the height ater seconds? c) What is a unction that describes the velocity o the rocket? d) What is the rocket s velocity ater seconds. e) What is a unction that describes the acceleration o the rocket? ) What is the rocket s acceleration ater 5 seconds? a) b) c) d) e) ) h 0 = 0 h = = 4 metres v t = h ' t = 9t v = 9 = 8 ms a t = v ' t = h '' t = 8t a 5 = 8 5 = 90 m s s A height o a ball is given by the unction; h t = t + t What is its maximum height? Max height will happen at a SP, thereore dierentiate; h ' t = t + SP happens where gradient =0; 0 = t + t = 8 thereore the max height happens ater 8 seconds h 8 = = 4 Thereore the Max height is 4 metres. Maths Quest Maths B Year or Queensland e

2 4 Find the stationary points o the unction = x x 4x 8 and state their nature. Also ind the intercepts with the axes and sketch the graph o the unction. SP s happen where slope is zero, hence need the gradient unction, and where the gradient is zero: x x x 8 ( x 4) ( x + ) = x x 4x 8 = x x 4 x = 4 and x = So, SP s are located where; x = and 4 Find nature o SP via second derivative '' x = x '' 4 = 4 = +ve Minimum '' 0 4 = ve Maximum Determine SP s: ( 4) = 4 ( 4) 4( 4) 8 = 08 ( ) = ( ) ( ) 4( ) 8 Thereore SP s are: (4, 08) and (, 0) *Long division is coming to Senior Maths B soon. Luckily you do not have to do this manually, but may use your calculator to do your actorisation J x = x x 4x 8 ( ) = ( ) ( ) 4( ) 8 Thereore x + is a actor x 5x 4 x + x x 4x 8 quotient : x x + x 5x 5x 4x 0x 4x 8 4x 8 5x 4 = 0 ( x 7)( x + ) So the unction becomes: x = (x + )(x + )(x 7) xintercepts are, and 7 yintercept is 8 Maths Quest Maths B Year or Queensland e

3 5 Sketch the graph o the unction, = x x + 9x, clearly showing all intercepts and turning points. *** PLEASE FOLLOW THE STRATEGIES IN THE SOLUTION TO THE PREVIOUS QUESTION. DO NOT FOLLOW THE SOLUTION ON THIS PAGE, BUT YOU CAN USE IT TO CHECK YOUR ANSWER. x 4x + x = and x = ( x )( x ) = x = x x + 9x x + 9 = + 9 = + 9 = 4 Stationary points are (, 0) and (, 4) = + 9 = \ ( 4) = ( 4) ( 4) + 9 = Thereore (, 0) is a minimum point. 0 = = = + 9 = \ Thereore (, 4) is a maximum point x = x x + 9x = x( x x + 9) = x( x ) xintercepts are 0 and yintercept is 0 Maths Quest Maths B Year or Queensland e

4 I = x + ax x + b, has a stationary point (, 0), ind the values o a and b. Find the other stationary point. = x + = x = + 4a a = b b = x + b + ax when + 4a + b = x x 4 ( x )( x + ) x = and x = The other stationary point at x = = + = ax Stationary point is (, ) x = 7 Find the stationary point and the nature o the stationary point or y = x x + x 8 y = x x + x 8 = x x + x x + x 4x + 4 when x = When x =, y = + 8 y '' = x y '' = Thereore, the point (,0) is a point o inlexion. Maths Quest Maths B Year or Queensland e 4

5 8 Find the equations o the tangent to the curve y = 5 x x at the points where the curve crosses the xaxis. *** Clearly you have spotted the book using that awul ormula. You will clearly NOT use this ormula to ind the equation o the line, but rather will use a point that lies on the line to ind c! *** and they have put the answer in a terrible orm it should be y = 8x + 4 ( 5 + x) y = 5 x x = Curve crosses xaxis at 5 and Gradient o tangent at x = 5 is = x = 5= 8 Equation o tangent at x = 5 is y 0 = 8 ( x 5) 0 = 8x y + 40 Gradient o tangent at x = is = x = = 8 Equation o tangent at x = is y 0 = 8 8x + y 4 ( ) ( x ) Maths Quest Maths B Year or Queensland e 5

6 9 Find the equations o the normal to the curve in the previous question at the points where y = 7 ** Same as last question Do not use that terrible ormula to ind the equation o a line. It does not show your understanding o how unctions work! x + x 8 ( x + 4)( x ) y = 5 x x = 7 x = 4 or x = Gradient o tangent at x = 4 is = x = 4= Gradient o normal is Equation o normal at x = 4 is ( x 4) y 7 = 0 = x + y 8 Gradient o tangent at x = is = x = = Gradient o normal is Equation o normal at x = is y 7 = x y + 40 ( x ) 0 Find the equation o the tangent and the equation o the normal to the curve: y = x at the point where x = 4. *** Why do they keep using that ormula? y = x = x = = =,at x= 4,and y= 4 Equation o tangent: ( x 4) y = x + y 5 Equation o normal: y = x y + 5 ( x 4) Maths Quest Maths B Year or Queensland e

7 Find the equation o the tangent to y = x x 4 which is parallel to the line x y +. *** again and again. Oh what are they teaching the youth o today L Find the equation o the tangent to y = x x which is perpendicular to the line x + y. *** this just makes me sad! x y +. y = x + Gradient = y = x x4 = x = x = 4 y = = y y y = x y 0 = m x x + y ( x ) ( x 4) y = x + y = x + Gradient o perpendicular line is. y = x x = x x = 5 x = 4 5 æ5ö 5 5 When x=, y = ç = 4 è4 ø 4 Equation o line is: y y = m xx 5 æ 5ö y = ç x è 4 ø 5 5 y = x 8 5 y = x Maths Quest Maths B Year or Queensland e 7

0,0 B 5,0 C 0, 4 3,5. y x. Recitation Worksheet 1A. 1. Plot these points in the xy plane: A

0,0 B 5,0 C 0, 4 3,5. y x. Recitation Worksheet 1A. 1. Plot these points in the xy plane: A Math 13 Recitation Worksheet 1A 1 Plot these points in the y plane: A 0,0 B 5,0 C 0, 4 D 3,5 Without using a calculator, sketch a graph o each o these in the y plane: A y B 3 Consider the unction a Evaluate