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13. (a) Express 10cosθ - 3sinθ in the form Rcos(θ + α), where R > 0 and 0 < α < 90º Give the exact value of R and give the value of α, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 2

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13.-(a)-Express-10cosθ---3sinθ-in-the-form-Rcos(θ-+-α),-where-R->-0-and-0-<-α-<-90º-Give-the-exact-value-of-R-and-give-the-value-of-α,-in-degrees,-to-2-decimal-places-Edexcel-A-Level Maths Pure-Question 15-2017-Paper 2.png

13. (a) Express 10cosθ - 3sinθ in the form Rcos(θ + α), where R > 0 and 0 < α < 90º Give the exact value of R and give the value of α, in degrees, to 2 decimal place... show full transcript

Worked Solution & Example Answer:13. (a) Express 10cosθ - 3sinθ in the form Rcos(θ + α), where R > 0 and 0 < α < 90º Give the exact value of R and give the value of α, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 15 - 2017 - Paper 2

Step 1

Express 10cosθ - 3sinθ in the form Rcos(θ + α)

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Answer

To express the given expression in the desired form, we start by identifying the constants involved:

  1. Set up the equation:

    R=extsqrt(102+32)=extsqrt(100+9)=extsqrt(109)R = ext{sqrt}(10^2 + 3^2) = ext{sqrt}(100 + 9) = ext{sqrt}(109)

  2. To find α, we use the tangent ratio:

    an(α)=310 an(α) = \frac{3}{10}

    To find α in degrees, calculate:

    α=an1(310)16.70ºα = an^{-1}\left(\frac{3}{10}\right) ≈ 16.70º

Thus, the values are:

  • R = √109
  • α = 16.70º

Step 2

find a complete equation for the model

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Answer

Given the initial height of 1 metre:

Using the earlier formula:

H=a10cos(80º)+3sin(80º)H = a - 10cos(80º) + 3sin(80º)

substituting for a:

  • Since the height is 1 metre at t = 0:
  • So, we have 1 = a - 10cos(80º) + 3sin(80º)
  • From here:
  • Solve to find a = 1 + 10cos(80º) - 3sin(80º).
  • Thus, complete equation is:

H=(1+10cos(80º)3sin(80º))10cos(80º)+3sin(80º)H = (1 + 10cos(80º) - 3sin(80º)) - 10cos(80º) + 3sin(80º)

Step 3

hence find the maximum height of the passenger above the ground

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Answer

The maximum height typically occurs at the peak points of the trigonometric components of the equation. To find the maximum value of H, we compute:

  1. Evaluate H when sin(80º) is maximized (at 1) and when cos(80º) is minimized (at -1):

    Hmax=a+10+3H_{max} = a + 10 + 3

  2. Substitute the value of a calculated previously:

    For maximum height, calculate: Hmax=(1+10cos(80º)3sin(80º))+10+3H_{max} = (1 + 10cos(80º) - 3sin(80º)) + 10 + 3

  3. Substitute numerical values to find H_max.

Step 4

Find the time taken, to the nearest second, for the passenger to reach the maximum height on the second cycle

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Answer

Using the periodic properties of H:

  1. Set while maximizing: 80t+(16.70º)=54080t + (16.70º) = 540

  2. Rearranging leads to: t=54016.7080t = \frac{540 - 16.70}{80}

  3. Compute value to find: t6.54t ≈ 6.54

  4. Thus, time rounded is approximately 6 minutes and 32 seconds.

Step 5

How would you adapt the equation of the model to reflect this increase in speed?

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Answer

To adapt for speed increment:

  1. Increase the coefficient '80' in the original equations in accordance with speed:

For example, use: H=a10cos(90º)+3sin(90º)H = a - 10cos(90º) + 3sin(90º) 2. The increase leads to a faster cycle time, thus affecting parameters accordingly.

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