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# Tutor profile: Rajasekar K.

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Rajasekar K.
Mathematics Expert
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## Questions

### Subject:Trigonometry

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Question:

The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds horizontally, the angle of elevation changes to 30°. If the aeroplane is flying at a velocity of 200m/s, then find the constant height at which the aeroplane is flying.

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Rajasekar K.

Let A be the point of observation. Let B and D be positions of the aeroplane initially and after 15 seconds respectively. Let's consider a triangle ABC with angleCAB = 60°, and triangle ADE with angleEAD = 30° AC=x metres, CE= 200*15= 3000metres BC=DE=h, In right-angled ▲AED, tan30° = h/(x+3000), h = (x + 3000) ×(1/√3) In right-angled ▲ACB, tan60° = h/x, h = x(√3) equating above two h's, we get 3x = x + 3000, x=1500m, then h=1500√3m height at which the aeroplane is flying is 2598 metres.

### Subject:Calculus

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Question:

To construct a cylindrical can with a bottom but no top that will have a volume of 30 cm3. Determine the dimensions of the can that will minimise the amount of material needed to construct the can.

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Rajasekar K.

V=πr²h, A=2πrh+πr² 30=πr²h h=30/πr² now, A(r)=2πr(30/πr²)+πr² = 60/r+πr² now, Differentiate A(r), A′(r)=−60/r² + 2πr = (2πr³−60)/r² now substitute A′(r)=0, to get critical value of r, r = ³√(60/2π) = 2.1216 now, Differentiate again A'(r), A′′(r)= 120/r³+2π now substitute A''(r)=0, to get max/min value of r, here second derivative is always positive, A(r)will always be concave up and so the single critical point must be a relative minimum. then h = 30/(π(2.1216)²) = 2.1215 solution: final dimensions r=2.1216cm, h=2.1215cm

### Subject:Differential Equations

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Question:

A 2 metre length metal rod has 1200 mm2 cross sectional area, with circumference are thermally insulated, one end connected with heat source of 10KW, another end maintained at 50oC, Determine the thermal distribution in the rod, thermal conductivity k=100KW/mC.

Inactive
Rajasekar K.

Heat flow per unit time (Q/t = qA = 10KW ), fourier law of heat conduction q(x) = -k dT(x)/ dx , Q = qA = -kA dT(x)/ dx, dT(x)/dx = -Q/KA = -10/((100)*1200*E-6) = -83.33 oC/m 1st order differential equation, T(x) = -83.33x + c, T(2) = 50oC, then c = 216.67 therefore temperature distribution is T (x) = 216.67 − 83.33x

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