12mm casio workout - weebly....%(2)=2×l×2+m=4l+m hence, slope of tangent is 4l+m. need to find the...

© John Wiley & Sons Australia, Ltd 2009 1

WorkSHEET CASIO WORKOUT Name: _____________________ 1 What is the gradient of 𝑦 = 2𝑥! − 3𝑥 − 1,

where 𝑥 = 2

𝑑𝑑𝑥(2𝑥! − 3𝑥 − 1)|"#! = 5

The slope where x=2 is 5.

2 In the test, if you are asked to differentiate;

𝑦 = 3𝑥! Your answer is

𝑦 = 6𝑥 How can you use your calculator to make sure you are correct?

Arbitrarily select 𝑥 = 1

𝑑𝑑𝑥(3𝑥!)|"#$ = 6

Check against

𝑓%(1) = 6 × 1 = 6 Check again, Arbitrarily select 𝑥 = 2

𝑑𝑑𝑥(3𝑥!)|"#! = 12

Check against

𝑓%(1) = 6 × 2 = 12 That’s good enough … you are correct!

3 The distance a particle is away from a given point is described by the function;

𝑑(𝑡) = 𝑥! + ln 𝑥 − sin 𝑥 What is the velocity of the particle after 2 seconds?

Correct, we don’t care that we have NO Idea what that function is. That’s what calculators are for!

We DO need to know that Velocity is the first

derivative of a distance function! Hence, the answer ;

𝑑𝑑𝑥(𝑥! + ln 𝑥 − sin 𝑥)|"#! = 4.916

The velocity of the object is 4.916 m/s

Maths Quest Maths B Year 11 for Queensland Chapter 13: Differentiation and applications WorkSHEET 13.1



𝑑(𝑡) = 𝑥! + ln 𝑥 − sin 𝑥 What is the acceleration of the particle after 5 seconds?

We also need to know that Acceleration is the

second derivative of a distance function!

𝑑!

𝑑𝑥!(𝑥! + ln 𝑥 − sin 𝑥)|"#& = 1.001

The acceleration of the particle is 1.001 m/s/s


𝑑(𝑡) = 𝑥! + ln 𝑥 − sin 𝑥 What is the Minimum Velocity of the particle?

Hmmm, how about we Graph the Velocity function, then just take a look and use G-Solve to find the minimum!

How do we graph a gradient function in the

Casio? Input the initial function

𝑌1 = 𝑥! + ln 𝑥 − sin 𝑥 Now, in Y2, input;

𝑌2 =𝑑𝑑𝑥(𝑌1)|"#"

De-Select Y1 so you don’t graph it, and only

graph Y2. Zoom – Auto G-Solve Min … 𝑥 = 0.622229𝑦 = 2.03902 ** Make sure your calculator is in RADIANS! Hence, Minimum Velocity is 2.039 at 0.622

seconds. ** you can also graph the gradient function as;

𝑌1 =𝑑𝑑𝑥

(𝑥! + ln 𝑥 − sin 𝑥)|"#" But then you do not have the option to look at

the distance graph!



6 The distance an object is away from a given point is described by the function;

𝑑(𝑡) =ln 𝑡𝑡'

What is the maximum acceleration of the particle?

Hmmm, how about we Graph the Acceleration function, then just take a look and use G-Solve to find the maximum!

How do we graph a gradient function in the

Casio? Input the initial function

𝑌1 =ln 𝑥𝑥'

Now, in Y2, input;

𝑌2 =𝑑!

𝑑𝑥!(𝑌1)|"#"

De-Select Y1 so you don’t graph it, and only

graph Y2. Remember that the second derivative of a

Distance / Time function is Acceleration, so when we graph (

!

("! we are looking at

a graph of acceleration, so we can “see” the minimum and maximum values, and can use G-Solve to find them!

V-Window – Standard G-Solve Max … 𝑥 = 2.188755974𝑦 =

0.0477776209 Hence, Maximum Acceleration is 0.0478 at

2.189 seconds. ** you can also graph the acceleration function

as;

𝑌1 =𝑑!

𝑑𝑥! >ln 𝑥𝑥' ?

|"#" But then you do not have the option to look at

the distance graph!

7 Solve for 𝑥;

𝑥! −1𝑥 − ln 𝑥 = 𝑥

𝑆𝑜𝑙𝑣𝑒𝑁 >𝑥! −1𝑥 − ln 𝑥 = 𝑥, 𝑥?

𝑥 = 1.6665



8 Determine the tangent to

𝑦 = 2𝑥! + 4𝑥 − 7 That has a slope of 8.

The tangent function is in the form,

𝑦 = 𝑚𝑥 + 𝑐

We know the slope of the tangent is 8, hence have;

𝑦 = 8𝑥 + 𝑐

Need to find the value of 𝑥 where the slope is

8. Casio;

𝑆𝑜𝑙𝑣𝑒𝑁 >𝑑𝑑𝑥(2𝑥! + 4𝑥 − 7)|"#" = 8, 𝑥?

𝑥 = 1

Therefore, the slope of the given function is 8,

where 𝑥 = 1 Find the actual point on the given function

where 𝑥 = 1

𝑓(1) = 2 × 1! + 4 × 1 − 7 = −1 So, (1, −1) is the Tangent point. Because this

point lies on the tangent, it must satisfy the tangent function, by sub;

−1 = 8 × 1 + 𝑐

𝑐 = −9 Hence, Tangent is;

𝑦 = 8𝑥 − 9

3



9 Determine the tangent to

𝑦 = 𝑥! − 4𝑥 − 7 Where 𝑥 = 1.

Need to find slope of the parabola where 𝑥 = 1 … Need gradient function;

𝑓%(𝑥) = 2𝑥 − 4

and 𝑓%(1) = 2 × 1 − 4 = −2

Hence, slope of Tangent is -2. *** if this is in the Tech Active, no need to do

this manually, just communicate use of GC:

𝑑𝑑𝑥(𝑥! − 4𝑥 − 7)|"#$ = −2

Need to find the Point on the Parabola, where

𝑥 = 1;

𝑓(1) = 1! − 4 × 1 − 7 = −10 Hence, Tangent has slope of -2 and passes

through the point (1, −10). Tangent function is in the form,

𝑦 = 𝑚𝑥 + 𝑐 and

𝑦 = −2𝑥 + 𝑐 As the Tangent goes through the point

(1, −10), it must satisfy the equation. By sub;

−10 = −2 × 1 + 𝑐

𝑐 = −8 Hence tangent is;

𝑦 = −2𝑥 − 8

10 How can you check your answer? Use your GC to graph;

𝑦 = 𝑥! − 4𝑥 − 7 and

𝑦 = −2𝑥 − 8 And use G-Solve to find POI. There is only on

and it happens where 𝑥 = 1, specifically (1, −10)



11 How ELSE can you check your answer? If these functions are tangential, there is only one POI, hence solve simultaneously;

𝑆𝑜𝑙𝑣𝑒𝑁(𝑥! − 4𝑥 − 7 = −2𝑥 − 8, 𝑥)

→ {1} There is only ONE solution, where 𝑥 = 1

12 Do we need to be able to do things different ways, using both technology and manually?

YES!

13 Lets do that last question again! Determine the tangent to

𝑦 = 𝐴𝑥! + 𝐵𝑥 + 𝐶 Where 𝑥 = 2.

Need to find slope of the parabola where 𝑥 = 2 … Need gradient function;

𝑓%(𝑥) = 2𝐴𝑥 + 𝐵

and 𝑓%(2) = 2 × 𝐴 × 2 + 𝐵 = 4𝐴 + 𝐵

Hence, slope of Tangent is 4𝐴 + 𝐵. Need to find the Point on the Parabola, where

𝑥 = 2; 𝑓(1) = 𝐴 × 2! + 𝐵 × 2 + 𝐶 = 4𝐴 + 2𝐵 + 𝐶

Hence, Tangent has slope of 4A and passes

through the point (2, 4𝐴 + 2𝐵 + 𝐶). Tangent function is in the form,

𝑦 = 𝑚𝑥 + 𝑐 and

𝑦 = (4𝐴 + 𝐵)𝑥 + 𝑐 As the Tangent goes through the point (2, 4𝐴 +

2𝐵 + 𝐶), it must satisfy the equation. By sub;

4𝐴 + 2𝐵 + 𝐶 = (4𝐴 + 𝐵) × 2 + 𝑐

𝑐 = −4𝐴 + 𝐶 Hence tangent is;

𝑦 = (4𝐴 + 𝐵)𝑥 − 4𝐴 + 𝐶 *** This is NOT difficult. It is the exact same process as the last tangent question, but now just tracking the pronumerals through carefully with your good algebra skills!



14 Where are the slopes of the functions of the following functions equal?

𝑓(𝑥) = 𝑥!

𝑔(𝑥) = 2(𝑥 − 1)!

Slope 𝑓(𝑥) = (("𝑥!

Slope 𝑔(𝑥) = (

("2(𝑥 − 1)!

Set slopes equal to each other and solve for the

value of 𝑥 that makes the slopes the same;

𝑆𝑜𝑙𝑣𝑒𝑁 '𝑑𝑑𝑥(𝑥!)|"#" =

𝑑𝑑𝑥(2(𝑥 − 1)!)|"#" , 𝑥2

𝑥 = 2

So, the slope of both functions is the same,

where 𝑥 = 2 Lets check this

𝑑𝑑𝑥(𝑥!)|"#! = 4

𝑑𝑑𝑥(2(𝑥 − 1)!)|"#! = 4

Therefore both functions have a slope of 4,

where 𝑥 = 2 Lets check this manually By inspection;

𝑓%(𝑥) = 2𝑥

𝑔%(𝑥) = 4𝑥 − 4 Set the slopes equal to each other

2𝑥 = 4𝑥 − 4

𝑥 = 2 Now,

𝑓%(2) = 2 × 2 = 4

𝑔%(2) = 4 × 2 − 4 = 4 That’s a mathematical way of calculating it, but

we could also graph them … and the slopes where 𝑥 = 2 would be the same … hence the tangents at these points of the graphs would be parallel! (this would justify our answer graphically?)

Done … Awesome!



15 Detailing the domain, range, axis intercepts, maxima/minima and asymptote details, sketch;

𝑦 =5 ln 𝑥𝑥 + 2

Take care … if a question is labelled as being worth 10 marks, do not miss anything!

Step 1: Graph it on your calculator. Step 2: Methodically answer ALL Questions: Domain: 0 < 𝑥 < ∞ Range: −∞ < 𝑦 < ∞ 𝑥-axis int: 𝑥 = 0.743 𝑦-axis int: No Intercept Maximum value: 3.839, at 𝑥 = 2.718 Minimum value: No minimum Asymptote: 𝑥 = 0 Asymptote: 𝑦 = 2 Then sketch it:

16 What is the difference between Solve and SolveN?

No idea … please let me know if you can work it out, but sometimes one is better than the other!



17 Solve for 𝑥. 𝑥! − 𝑥 − 6 = 0

Casio: Option F4 – Calculate F1 – Solve

𝑆𝑜𝑙𝑣𝑒(𝑥! − 𝑥 − 6 = 0, 𝑥) Answer = 3 So Solve is a DUD here … rather use Solve N Casio: Option F4 – Calculate F5 – SolveN

𝑆𝑜𝑙𝑣𝑒𝑁(𝑥! − 𝑥 − 6 Answer = - 2 or 3 Yeehaa … ! Note … you do not need to finish the input, the calculator assumes anything to the right of where you finish is = zero and it solves of x automatically! So here, SolveN works, where Solve is NOT a good option!

18 Use technology to help you Graph,

𝑦 = 𝑥! − 2𝑥 − 8

Make sure you can input the function to graph it. Then use G-Solve to find: Roots; 𝑥 = −2, 4 Min; (1, −9) y-axis int; 𝑦 = −8 and sketch … LABELLING these important points!



19 Do NOT use technology to help you Graph,

𝑦 = 𝑥! − 2𝑥 − 8

Factorise to find x-axis intercepts Set x=0 to find y-axis intercepts Last year we would have completed the square to find the TP, but now we differentiate;

𝑦% = 2𝑥 − 2 SP happen where slope -0

0 = 2𝑥 − 2

𝑥 = 1

𝑦(1) = 1 − 2 − 8 = −9

𝑇𝑃(1,−9) No, you won’t just need to graph Parabolas anymore … there can be cubics and higher!

20 This one could have been a Tech Active question from last term. It has nothing to do with differentiation or probability, it is just about you knowing how to use technology.

The altitude, in metres, with respect to time, in seconds, of a roller-coaster is modelled by the function:

ℎ(𝑡) = −𝑒$)*"#.%& *+ ,- . + 15

where 0 < 𝑡 ≤ 30

Validate the claim by the theme park’s engineers stating that “75% of the time the rollercoaster is at least 10m above the ground”.

Verify whether this statement is true.

Graph the function:

𝑦 = −𝑒$)*'/.&" *+ ,- " + 15

Find where the height is above 10 … ?

Graph the function:

𝑦 = 10

Adjust window to see clearly;

−1 < 𝑥 < 31

−1 < 𝑦 < 15

G-Solve to find intercepts:

𝑥 = 3.4848𝑎𝑛𝑑10.4868

From the domain restriction in the question, the ride lasts for 30 seconds.

The ride is above 10 metres between

0 to 3.4848 and 10.4868 to 30

Total time above 10 metres is 22.998

As a percentage

22.99830 × 100 = 76.66%

Therefore the engineers claim is valid!



21 This one could have been a Tech Active question from last term. It has nothing to do with differentiation or probability, it is just about you knowing how to use technology.

The altitude, in metres, with respect to time, in seconds, of a roller-coaster is modelled by the function:

ℎ(𝑡) = −𝑒$)*"#.%& *+ ,- . + 15

where 0 < 𝑡 ≤ 30

Validate the claim by the theme park’s engineers stating that “75% of the time the rollercoaster is at least 10m above the ground”.

Verify whether this statement is true.

Alternate solution; Set ℎ = 10 𝑆𝑜𝑙𝑣𝑒𝑁(10 = −𝑒$)*

'/.&" *+ ,- " + 15, 𝑥)

𝑥 = 3.4848𝑎𝑛𝑑10.4868

Verify where “above 10” is located;

𝑓(3) = (ℎ𝑖𝑔ℎ𝑒𝑟𝑡ℎ𝑎𝑛10)

𝑓(4) = (𝑙𝑒𝑠𝑠𝑡ℎ𝑎𝑛10)

𝑓(10) = (𝑙𝑒𝑠𝑠𝑡ℎ𝑎𝑛10)

𝑓(11) = (ℎ𝑖𝑔ℎ𝑒𝑟𝑡ℎ𝑎𝑛10)

Also note domain restriction end time of ride is 30 seconds The ride is above 10 metres between

0 to 3.4848 and 10.4868 to 30

Total time above 10 metres is 22.998

As a percentage

22.99830 × 100 = 76.66%

Therefore the engineers claim is valid!



22 Determine the area bounded by the functions;

𝑓(𝑥) = −𝑥!

and

𝑔(𝑥) = 𝑥! − 𝑥 − 6

Casio, Step 1. Graph both functions; Step 2. Shift F5 (G-Solve) Step 3. F6 (arrow right) Step 4. F3 (integrate symbol) Step 5. F3 (intersect) Step 4. The calculator automatically finds the left most point of intersection, hit “Execute” to select that POI Step 5: move right (press on the right side of the solver navigation button) Step 6. Hit “Execute” to select that next POI Step 6. Write down your answer.


𝑓(𝑥) = −𝑥!

And

𝑔(𝑥) = 𝑥! − 𝑥 − 6

As it would appear in your exam; Use G-Solve to find area between POI’s; Area = 14.292𝑢! Do a quick rough sketch!



24 WithOUT using G-Solve, determine the area bounded by the functions;

𝑓(𝑥) = −𝑥!

and

𝑔(𝑥) = 𝑥! − 𝑥 − 6

We don’t need G-Solve: Find POI’s

𝑆𝑜𝑙𝑣𝑒𝑁(−𝑥! = 𝑥! − 𝑥 − 6)

𝑥 = −32 , 2

Use Area between is Area top – area Bottom … ∫ 𝑓(𝑡𝑜𝑝) − 𝑓(𝑏𝑜𝑡𝑡𝑜𝑚)12 ** Clearly 𝑓(𝑥) = −𝑥! is on top (Do a quick rough sketch!)

` −𝑥! − (𝑥! − 𝑥 − 6)!

*'!

Area = 14.292𝑢! ** did you see the need to communicate the use of technology? ** did you see the need to Justify which function was on top? (Yes, there are MARKS everywhere, but it doesn’t take a paragraph to communicate or justify)




𝑓(𝑥) = −𝑥! − 𝑥 − 1

and

𝑔(𝑥) = 𝑥! − 3𝑥 − 13

Evaluate the reasonableness of your solution by using an alternative method.

Do a quick rough sketch … ! Use G-Solve to find area between POI’s; Area = 41.67𝑢! Alternate solution; Find POI’s

𝑆𝑜𝑙𝑣𝑒𝑁(−𝑥! − 𝑥 − 1 = 𝑥! − 3𝑥 − 13) 𝑥 = −2, 3

Use Area between is Area top – area Bottom … ∫ 𝑓(𝑡𝑜𝑝) − 𝑓(𝑏𝑜𝑡𝑡𝑜𝑚)12 ** Clearly 𝑓(𝑥) = −𝑥! − 𝑥 − 1 is on top

` −𝑥! − 𝑥 − 1 − (𝑥! − 3𝑥 − 13)'

*!

Area = 41.67𝑢!



26 The derivative of a function is

𝑓%(𝑥) =2√𝑥

−𝑘𝑥

It has a stationary point where 𝑥 = 1 and it passes through the point (5, 8.725).

Determine the value of 𝑘 and determine the original function.

Justify the reasonableness of your solution with a sketch.

SP àslope is zero where 𝑥 = 1

𝑆𝑜𝑙𝑣𝑒𝑁 >0 =2√1

−𝑘1 , 𝑘?

𝑘 = 2

𝑓%(𝑥) =2√𝑥

−2𝑥

Now integrate to find 𝑓(𝑥)

𝑓(𝑥) = 4√𝑥 − 2 ln 𝑥 + 𝑐

Find c … (1,5) lies on the line, so it must satisfy the eqn. by sub;

𝑆𝑜𝑙𝑣𝑒𝑁c8.725 = 4√5 − 2 ln 5 + 𝑐, 𝑐d

𝑐 = 3 *** note Casio gives 2.999603915 … =3 J

𝑓(𝑥) = 4√𝑥 − 2 ln 𝑥 + 3



27 This is an old “B” level exam question.

A block of land is represented below. It is bounded by the 𝑥 and 𝑦 axes and the functions;

𝑦 = +33"

, 𝑦 = 50 and 𝑥 = 60

The owners wish to subdivide this block into two regions of equal area, but are unsure where to place the dividing boundary.

You are required to appropriately position the proposed subdivision boundary in the form of the function 𝑥 =?

Find POI;

𝐺𝑠𝑜𝑙𝑣𝑒 → (12,50)𝑎𝑛𝑑(60,10) Think … total area is;

` 50$!

3𝑑𝑥 +`

600𝑥

+3

$!

Need to split that at an unknown point … 𝑥 = 𝑘

How about

𝐴𝑟𝑒𝑎𝐴 = 𝐴𝑟𝑒𝑎𝐵

` 50$!

3𝑑𝑥 + `

600𝑥

4

$!= `

600𝑥

+3

4

So,

𝑆𝑜𝑙𝑣𝑒𝑁 56 50$!

%𝑑𝑥 +6 '

600𝑥 2

&

$!= 6 '

600𝑥 2

'%

&, 𝑘<

𝑘 = 16.27492543

Therefore, the boundary should be placed where k = 16.275



28 The person in that last question changed their mind and wants their dividing fence to go horizontally.

Can you re-do that solution?

Do a rough sketch;

Need to Integrate between bounds of 𝑥, but don’t know 𝑦 =? Set;

𝑦 = 𝑐 Find POI between two functions,

𝑦 = +33"

and 𝑦 = 𝑐

600𝑥 = 𝑐

𝑥 =600𝑐

Therefore POI is … g+33

5, 𝑐h

Hence, 𝑘 = +33

5

Establish Integral Equation in Casio; (too big for half a page … )

29 𝐴𝑟𝑒𝑎𝐴 = 𝐴𝑟𝑒𝑎𝐵

𝑆𝑜𝑙𝑣𝑒𝑁i` (50 − 𝑐)$!

3𝑑𝑥 +` >

600𝑥 − 𝑐?

+335

$!𝑑𝑥 = ` 𝑐

+335

3𝑑𝑥 +` >

600𝑥 ?

+3

+335

𝑑𝑥, 𝑐j

𝑐 = 13.56244

Therefore, the boundary should be placed where 𝑦 = 13.56244 (I’ll leave it to you to check it!)



30 The area between the function𝑦 = 𝑥', the positive 𝑥-axis and the line 𝑥 = 𝑘 is 64. Determine 𝑘.

Area is given through Integration; Use GC;

𝑆𝑜𝑙𝑣𝑒𝑁 k` 𝑥'4

3𝑑𝑥 = 64, 𝑘l

∴ 𝑘 = 4

31 Determine the area under the function;

𝑦 = 𝑥' − 2𝑥! − 5𝑥 + 6

for the domain;

0 < 𝑥 < 4.

` 𝑥' − 2𝑥! − 5𝑥 + 6/

3𝑑𝑥

Area = $+

'𝑢!

*** Note, even though some of this area is under the x-axis, the calculator is smart enough just to work out the net area!

32 This time use your G-Solve to determine the area under the function;

𝑦 = 𝑥' − 2𝑥! − 5𝑥 + 6

for 0 < 𝑥 < 4.

Graph the function, use g-solve and select bounds of

𝑥 = 0𝑎𝑛𝑑4 Area = 5.3333333

33 Hmmm … are you sure about your last 2 answers … ?

Use GSolve to determine the area under the function

𝑦 = 𝑥 − 1

for 0 < 𝑥 < 2.

Oh darn … ! How can that area = Zero? Yes, even in G-Solve, the calculator is not that smart! So lets go back two questions … refer next question

34 Determine the area under the function;

𝑦 = 𝑥' − 2𝑥! − 5𝑥 + 6

for the domain;

0 < 𝑥 < 4.

G-Solve is the BEST way of doing this, but you need to identify the function cuts the x-axis. F5 – Gsolve F6 – move right F3 - ∫ 𝑑𝑥 F4 – Mixed 𝑋, 𝜃, 𝑇 – 0 – Execute 𝑋, 𝜃, 𝑇 – 4 – Execute Although it gives the wrong answer against ∫𝑑𝑥 = 5.3333333, it also gives the correct answer against a shaded box Area = 16

12mm casio workout - weebly....%(2)=2×l×2+m=4l+m hence, slope of tangent is 4l+m. need to find the...

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