QCE Specialist Maths: Differential equations and related rates Q256

Q256 · Practice questionTechnology-activeVery complex unfamiliar12 marks

QUESTION 256 (12 marks)

In a robot simulation, energy

E

joules obeys

E^{\prime}=-kE-v^2

, where

k=\ln2/5

^{-1}

and

v

is the numerical speed in m/s. The coefficient of the

v^2

term is included in the simulation units. Initially

E=420

. The robot travels for 5 s at constant speed

u

, then for 5 s at constant speed

w

, with

u,w\ge0

. It must travel exactly 30 m. A subsequent task requires independent exponential energy

X

with mean 20 J. Success means

X\le E(10)

Derive

E(10)

as a function of

u,w

, using the energy at the end of the first stage as the initial condition for the second.

[3 marks]

Determine the two speeds that maximise the remaining energy, and give the maximum energy to three decimal places.

[3 marks]

Determine the greatest achievable success probability for the subsequent task, to four decimal places.

[2 marks]

Determine the full interval of first-stage speeds

u

for which success probability is at least

0.95

. Give its endpoints to three decimal places and verify that energy remains positive throughout each admissible route.

[4 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

E(10)=105-\frac{5}{2\ln2}\left(\frac{u^2}{2}+w^2\right)

. (b)

u=4

w=2

m/s;

E_{\max}=105-30/\ln2=61.719

J. (c)

P_{\max}=1-e^{-(105-30/\ln2)/20}=0.9543

. (d)

3.422\le u\le4.578

m/s (rounded endpoints).

Worked solution

(a) At fixed speed

v

E(t)=(E(0)+v^2/k)e^{-kt}-v^2/k

. Since

e^{-5k}=1/2

, the first stage gives

E(5)=210-u^2/(2k)

. The second gives

E(10)=E(5)/2-w^2/(2k)=105-u^2/(4k)-w^2/(2k)

, equivalent to the stated form.

(b) The distance constraint is

5u+5w=30

, so

w=6-u

and

0\le u\le6

. The weighted cost is

C(u)=u^2/2+(6-u)^2=\tfrac32(u-4)^2+12

. It has a unique minimum at

u=4

, giving

w=2

and

E_{\max}=105-30/\ln2

e

P(X\le e)=1-e^{-e/20}

, an increasing function of

e

. Therefore the energy-maximising speeds also maximise success, giving

0.9543137\ldots

(d) The probability condition requires

E(10)\ge20\ln20

. Thus

\tfrac32(u-4)^2+12\le B

, where

B=\frac{2\ln2}{5}(105-20\ln20)

. The exact interval is

4-\sqrt{\frac23(B-12)}\le u\le4+\sqrt{\frac23(B-12)}

, with endpoints

3.422468\ldots

and

4.577532\ldots

. These lie within

[0,6]

and give nonnegative

w=6-u

. Final energy is at least

20\ln20>0

. While

E\ge0

, its derivative is negative at either positive stage speed; the fixed-speed solution therefore decreases to its positive endpoint on each stage. It cannot cross zero and later recover, so the energy stays positive throughout.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Solve the linear energy equation for a constant speed.

[1 mark]

Part a: Apply the first-stage duration and initial energy.

[1 mark]

Part a: Use continuity at the stage change to obtain the final energy.

[1 mark]

Part b: Use the distance constraint and physical speed domain.

[1 mark]

Part b: Minimise the weighted quadratic and justify a global maximum of energy.

[1 mark]

Part b: Give both speeds and the maximum energy.

[1 mark]

Part c: Use the exponential distribution and explain the monotonic optimisation.

[1 mark]

Part c: Calculate the maximal success probability.

[1 mark]

Part d: Translate the probability target into the energy threshold.

[1 mark]

Part d: Form and solve the quadratic inequality.

[1 mark]

Part d: Give the whole physical interval at the requested accuracy.

[1 mark]

Part d: Justify positive energy throughout the route.

[1 mark]

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Differential equations and related rates — Question 256

QUESTION 256 (12 marks)

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