Q.381. Angle between tangents drawn from a point to a parabola.

Question 381.
What is the angle between the tangents drawn from the point (1,4) to the parabola y^2 = 4x ?

Answer 381.
y = mx + a/m is the tangent to the parabola y^2 = 4ax
For y^2 = 4x, a = 1
=> y = mx + 1/m is the tangent

If it passes through (1, 4),
4 = m + 1/m
=> m^2 - 4m + 1 = 0
The roots m1 and m2 are the slopes of the tangents
=> m1 + m2 = 4 and m1m2 = 1
=> m1 - m2
= √[(m1 + m2)^2 - 4m1m2]
= √(16 - 4)
= 2√3

If θ = angle between the tangents, then
tanθ
= l (m1 - m2) / (1 + m1m2) l
= l 2√3 / (1 + 1) l
= √3
=> θ = arctan√3 = π/3.

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Q.380. Centroid of a triangle.

Question 380.
Segments AD = 10, BE = 6 and CF = 24 are drawn from the vertices of triangle ABC , each perpendicular to a straight line RS, not intersecting the triangle. Points D, E and F are the intersection points of RS with the perpendiculars. If x is the length of the perpendicular segment, GH, drawn to RS from the intersection point, G, of the medians of the triangle, then find the value of x.

Answer 380.
Treat RS as x-axis.
=> y-coordinate of A, B and C are
AD=10, BE = 6 and CF = 24 respectively.
=> y-coordinate of the median G
= GH = (1/3) (10 + 6 + 24) = 40/3.
=> x = 40/3.

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Q.379. Refraction of light.

Question 379.
You need to hit an underwater target lying flat on the bottom of a pool. The water is 1 m deep and you are standing so that your eyes are 3 m above the bottom of the pool. As you look at the target your gaze is 30 degrees below the horizontal. At what angle below the horizontal should you throw a spear to hit the target? (assume it travels in a straight line and that you throw at the same level as your eyes).

Answer 379.
Refer to the figure shown below.

For refraction,
sin i / sin r = 4/3
=> sin r = (3/4) sin(60°) = 3√3/8
=> tan r = tan [arcsin (3√3/8)]
=> tan r = 0.8542421962
=> AB = 0.8542421962

MB = 2cot (60°) = 3.4641016151

=> MA
= MB + BA
= 0.8542421962 + 3.4641016151
= 4.31834381

=> required angle,
θ = arctan (3/4.31834381) = 34.79°.

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Q.378. Trigonometric derivation.

Question 378.
Express tan5A in terms of tanA.

Answer 378.
tan5A
= tan(2A + 3A)
= (tan2A + tan3A) / (1 - tan2A tan3A) ... ( 1 )

tan2A + tan3A
= 2tanA / (1 - tan^2 A) + (3tanA - tan^3 A) / (1 - 3tan^2 A)
= (2tanA - 6tan^3 A + 3tanA - 3tan^3 A - tan^3 A + tan^5 A)
    diided by [(1 - tan^2 A)(1 - 3tan^2A]
= (tan^5 A - 10tan^3 A + 5tanA) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 2 )

1 - tan2A tan3A
= 1 - [2tanA (3tanA - tan^3 A)] / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 4tan^2 A + 3tan^4 A - 6tan^2 A + 2tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A]
= (1 - 10tan^2 A + 5tan^4 A) / [(1 - tan^2 A)(1 - 3tan^2A] ... ( 3 )

Putting results ( 2 ) and ( 3 ) in ( 1 ),
tan5A
= (tan^5 A - 10tan^3 A + 5tanA) / (1 - 10tan^2 A + 5tan^4 A).

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Q.377. Indefinite integration.

Question 377.
Find ∫ [(1 + sinx) / (1 + cosx)] e^x dx.

Answer 377.
e^x * [(1 + sinx) / (1 + cosx)]
= e^x * [1 + 2sin(x/2) cos(x/2)] / [2cos^2 (x/2)]
= [(1/2) sec^ (x/2) + tan(x/2)] e^x

 Let f (x) = tan(x/2)
=> f '(x) = (1/2) sec^2 (x/2)

 => ∫ e^x [(1+sinx)/(1 + cosx)] dx
= ∫ [(1/2) sec^ (x/2) + tan(x/2)] e^x dx
= ∫ [ f (x) + f '(x) ] e^x dx
= f (x) e^x + c
= tan(x/2) e^x + c.

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Q.376. Motion in two dimensions.

Question 376.
The velocity of a particle is v = {3i + (6 - 2t)j} m/s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s.

Answer 376.
v = dr/dt = 3i + (6 - 2t)j
=> r = 3t i + (6t - t^2) j + c
t = 0 => r = 0 => c = 0
=> r_t = 3t i + (6t - t^2) j
r_3 = 9i + 9j
r_1 = 3i + 5j
=> displacement during the time interval t = 1 s to t = 3 s
= r_3 - r_1
= 6i + 4j
=> magnitude of the displacement
= √(6^2 + 4^2)
= 2√(13) m.
Direction is given by
arctan(2/3) anticlockwise with x-axis
= 71.57° anticlockwise with x-axis.

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Q.375. First law of thermodynamics.

Question 375.
Initially 0.800 mol of an ideal gas in a container occupies a volume of 3.10 l at a pressure of 3.90 atm with an internal energy U1 = 364.8 J. The gas is cooled at a constant volume until its pressure is 2.50 atm. Then it is allowed to expand at constant pressure until its volume is 6.20 l. The final internal energy is U2 = 467.7 J. All processes are quasi static. Draw this process on a PV diagram. What is the work done by the gas? What is the heat absorbed by the gas?

Answer 375.
The processes are shown on PV diagram as under.

Work done by the gas in isochoric process = 0
Work done bt the gas in isobaric process = PdV
= (2.50) * (6.20 - 3.10) lit-atm
= 7.75 lit-atm
= 785.26875 joule ... [Refer to the link - http://www.convertunits.com/from/L+*+atm…

Heat absorbed by the gas = Increase in internal energy + work done by the gas
= 467.7 - 364.8 + 785.3 joule
= 888.2 joule
= (888.2) * (0.239) calories
= 212.3 calories.

[No. of moles of the gas is not needed in any of the above calculations.]

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Q.374. Center of mass

Question 374.
The assembly is made from a steel hemisphere, ρ_st= 7.80 Mg/m³, and an aluminum cylinder, ρ_al= 2.70 Mg/m³. Determine the height h of the cylinder so that the mass center of the assembly is located at z(bar)= 160 mm.

Answer 374.

Mass of steel hemisphere, m1

= (2/3) π r1^3 * ρ_st
= (2/3) π (16)^3 * (7.8) g
= 66913 g

Mass of aluminium cylinder, m2
= π r2^2 h * ρ_al
= π (8^2) h * 2.7 g
= 543h g

 => 16 = [m1 * (5/8) * 16 + m2 * (16 + h/2)] / (m1 + m2)
=> 16m1 + 16m2 = 10m1 + 16m2 + (m2/2) h
=> (m2/2) h = 6m1
=> (543/2) h^2 = 6 * 66913
=> h^2 = (12 * 66913) / 543
=> h = 38.45 cm = 384.5 mm.

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Q.373. Center of Mass.

Question 373.
Determine the distance h to which a hole must be bored into the cylinder so that the center of mass of the assembly is located at
x = 64 mm.

Answer 373.
As the cylinder is of uniform density, the center of mass = center of volume.

Volume of cylinder before drilling,
V1 = π * 4^2 * 12 = 192π cc

Volume of drilled hole,
V2 = π * 2^2 * h = 4πh cc

Center of mass = center of volume
=> 6.4 = (V1*6 - V2*(h/2) / (V1 - V2)
=> 6.4 = [192π * 6 - 4πh * (h/2)] / (192π - 4πh)
=> 6.4 * (192 - 4h) = (1152 - 2h^2)
=> 3.2 * (192 - 4h) = (576 - h^2)
=> h^2 - 12.8h + 38.4 = 0
=> (h - 8) (h - 4.8) = 0
=> h = 4.8 cm or 8 cm.

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Q.372. Projectile Motion.

Question 372.
The ball is thrown off the top of the building. If it strikes the ground at B in 3 s, determine the initial velocity v_a and the inclination angle θ_a at which it was thrown. Also, find the magnitude of the ball's velocity when it strikes  the ground.

Answer 372.
Horizontal and vertical components of the velocity are
v_a cos(θ_a) and v_a sin(θ_a) respectively.

For the horizontal displacement,
v_a cos(θ_a) * 3 = 60
=> v_a cos(θ_a) = 20 ... ( 1 )

For the vertical displacement,
v_a sin(θ_a) * 3 - (1/2) g*(3^2) = - 75
=> v_a sin(θ_a) = 23 ... ( 2 )

Squarring and adding eqns. ( 1 ) and ( 2 ),
v_a^2 = 20^2 + 23^2
=> v_a = 30.5 ft/s

Plugging this value in eqn. ( 1 ),
cos(θ_a) = (20)/(30.5)
=> θ_a = 49°.

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Q.371. Vectors application

Question 371.
Refer to the following figure.

In triangle ABC with two inner line-segments AD and CE, 
BD : DC = 3:1 and AE : EB = 2:1.
Find the ratio of CG : GE and AG : GD .

Answer 371.
Let a and c be the position vectors of A and C with respect to B as null vector.
=> Position vector of D = 3c/4 and that of E = a/3

Let AG : GD = m : 1
=> position vector of G is 1/(m + 1) * (3mc/4 + a) ... ( 1 )

Let EG : CG = n : 1
=> position vector of G is 1/(n + 1) * (nc + a/3) ... ( 2 )

Comparing coefficients of a in ( 1 ) and ( 2 ),
m + 1 = 3(n + 1) => m = 3n + 2 ... ( 3 )
Comparing coefficients of c in ( 1 ) and ( 2 ),
3m/[4(m + 1)] = n/(n + 1) ... ( 4 )

Plugging m = 3n + 2 from ( 3 ) in ( 4 ),
3 (3n + 2)/[4(3n + 3) = n/(n + 1)
=> 3n + 2 = 4n
=> n = 2
and from ( 3 ),
m = 3n + 2 = 8

=> AG : GD = m : 1 = 8 : 1
and EG : CG = n : 1 = 2 : 1.

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Q.370. First law of thermodynamics - work done problem.

Question 370.
A cylinder/piston setup contains air at 100kPa and 20° C and has a volume of  0.3 m^3. The air is compressed to 800 kPa in a reversible process in which PV^1.2 is held constant, after which it is expanded back to 100 kPa in a reversible adiabatic process.
Calculate the final temperature and the net work.

Answer 170.
Given equation PV^1.2
=> P1V1^1.2 = P2V2^1.2 ... ( 1 )

According to Ideal Gas Law equation,
PV = nRT
=> P1V1/T1 = P2V2/T2
=> P1V1T2 = P2V2T1
=> (P1V1T2)^1.2 = (P2V2T1)^1.2 ... ( 2 )

From ( 1 ) and ( 2 ),
P1^0.2 * T2^1.2 = P2^0.2 * T1^1.2
=> T2 = T1 * (P2/P1)^0.2/(1.2)
=> Final temperature,
T2 = (20 + 273) * (8)^(1/6) = 414.4 K = 141.4° C

Work done, W
= RT1 / (n - 1) * [1 - (p2/p1)^(n-1)/n]
= (8.314) * (293) * [1 - (8)^1/6)
= - 1009 J

Work done in adiabatic expansion, W'
= RT2 / (γ - 1) * [1 - (p2/p1)^(γ-1)/γ] ... γ = 1.41 for air
= (8.314) * (414.4) * [1 - (1/8)^(0.41/1.41)]
= 1563 J

Net work done
= - 1009 + 1563 J
= 554 J.

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Q.369. Integration.

Question 369.
Evaluate  ∫ dx / (x^2 + 2x + 2)^2.

Answer 369.
x^2 + 2x + 2
= (x + 1)^2 + 1

Let x + 1 = tanu
=> (x + 1)^2 + 1 = tan^2 u + 1 = sec^2 u
and dx = sec^2 u du

=> Integral
= ∫ sec^2 u du / sec^4 u
= (1/2) ∫ 2cos^2 u du
= (1/2) ∫ (1 + cos2u) du
= (1/2) u + (1/4)sin2u + c
= (1/2) u + (1/2) tanu / (1 + tan^2 u) + c
= (1/2) arctan (x + 1) + (1/2) (x + 1) / [(x + 1)^2 + 1] + c
= (1/2) arctan (x + 1) + (1/2)(x + 1)/(x^2 + 2x + 2) + c
= (1/2) [arctan(x + 1) + (x + 1)/(x^2 + 2x + 2)] + c.

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Q.368. Maximising area of a trapezoid.

Question 368. Find the maximum area of a trapezoid whose area is given by
A = 4sin(x)(16+16cos(x)) = 64sin(x) + 64sin(x)cos(x).

Answer 368.
For A to be maximum, dA/dx = 0 and d^2A/dx^2 < 0
dA/dx = 0
=> 64cosx + 64cos2x = 0
=> cos2x + cosx = 0
=> 2cos^2 x + cosx - 1 = 0
=> 2cos^2 x + 2cosx - cosx - 1 = 0
=> (2cosx - 1) (cosx + 1) = 0
=> cosx = 1/2 or - 1
=> x = 60° or 270°
d^2A/dx^2 = - 64sinx - 128sin2x < 0 for x = 60°
and maximum area
=A (max)
= 64sin60° + 32sin120°
= 96sin60°
= 83.14 sq. units.

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Q.367. To find coefficients of a quadratic equation given a condition.

Question 367.
Using y = Ax^2 + Bx + C,  how would you solve for y'' - 4y' + 3y = 9x^2 ?

Answer 367.
y = Ax^2 + Bx + C
=> y' = 2Ax + B
=> y" = 2A
=> y" - 4y' + 3y = 9x^2
=> 2A - 4 (2Ax + B) + 3(Ax^2 + Bx + C) = 9x^2
=> 3Ax^2 + (3B - 8A)x + 2A - 4B + 3C = 9x^2
Comparing coefficients of x^2, 3A = 9 => A = 3,
Comparing coefficients of x, 3B - 8A = 0 => B = (8/3)A = 8 and
comparing the constant terms, 2A - 4B + 3C = 0 => C = (1/3) (4B - 2A) = 26/3
Answers:
A = 3,
B = 8 and
C = 26/3
================================
Verification:
y = 3x^2 + 8x + 26/3
=> y' = 6x + 8
=> y" = 6
=> y''- 4y' + 3y
= 6 - 4 (6x + 8) + 3 (3x^2 + 8x + 26/3)
= 6 - 24x - 32 + 9x^2 + 24x + 26
= 9x^2 (confirmed).

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Q.366. Compound interest - equatted monthly intalments of loan

Question 366.
A woman borrows $50 000 in order to buy a house. Compound interest at the rate of 12% per annum in charge on the loan. She agrees to pay back the loan in 25 equal instalments at yearly intervals, the first repayment being made exactly one year after the loan is taken out. Calculate the value of each instalment.

Answer 366.
Let the equal yearly instalments = $ x

Loan outstanding at the end of year 1 = $ 50000 * (1.12) - x
Loan outstanding at the end of year 2
= $ [50000 * (1.12) - x]*(1.12) - x
= $ 50000 * (1.12)^2 - x * (1 + 1.12)

Loan outstanding at the end of year 3
= $ [50000 * (1.12)^2 - x * (1 + 1.12)] * (1.12) - x
= $ 50000 * (1.12)^3 - x [1 + 1.12 + (1.12)^2]

Loan outstanding at the end of year 25
= $ 50000 * (1.12)^25 - x [1 + (1.12) + ... + (1.12)^24]

Loan outstanding at the end of year 25 should be zero
=> x * [1 + 1.12 + (1.12)^2 + ... + (1.12)^24] = 50000 * (1.12)^25
=> x * [(1.12)^25 - 1] / (1.12 - 1) = 850003.220332
=> x * (17.00006 - 1) = (0.12) * (850003.220332)
=> 16.00006 x = 102000.386
=> x = 6375
=> yearly equal instalments
= $ 6375 each year.

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Q.365. Rectilinear motion.

Question 365.
At time t = 0 a car has a velocity of 16 m/s. It slows down with an acceleration given by -0.50t, in m/s^2 for t in seconds. What is the distance travelled by the car by the time it stops?

Answer 365.
dv/dt = - 0.5 t
=> dv = - 0.5 t dt

Integrating,
v = - (1/4) t^2 + c
t = 0 => v = 16 => c = 16
=> v = - (1/4) t^2 + 16 ... ( 1 )
=> ds/dt = - (1/4) t^2 + 16
=> ds = [16 - (1/4) t^2] dt

Integrating,
s = 16t - (1/12) t^3 + c
t = 0 => s = 0 => c = 0
=> s = 16t - (1/12)t^3 ... ( 2 )
When it stops v = 0
=> from eqn. ( 1 ), 0 = - (1/4) t^2 + 16 => t = 8
Plugging this value of t in eqn. ( 2 ),
distance travelled, s 
= 16 * 8 - (1/12) * 8*3
= 128 - 42.7
= 85.3 m.

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Q.364. Number of tangents to a curve and points of contacts.

Question 364.
How many tangent lines to the curve y= x/(x + 1) pass through the point (1,2)? At which points do these tangent lines touch the curve?

Answer 364.
The line through (1, 2) having slope m is
y - 2 = m (x - 1)

The tangent to a curve is a line which intersects the curve in two or more identical points. Hence, solving the above line with the curve,
y = x/(x + 1) = 2 + m(x - 1)
=> x = 2(x + 1) + m (x^2 - 1)
=> mx^2 + x + 2 - m = 0
The roots of this quadratic eqn. must be identical
=> its discriminant = 0 and the equal roots are x = 1/(2m)
=> 1 - 4m(2 - m) = 0
=> 4m^2 - 8m + 1 = 0
=> m^2 - 2m + 1/4 = 0
=> (m - 1)^2 = 3/4
=> m - 1 = ±√3/2
=> m = (1/2) (2 ± √3)
Two values of m indicate that there are two tangent lines.

The points of contact are given by
x = - 1/2m = - (2 - √3) or - (2 + √3)
For x = - (2 - √3),
y = x/(x+1) = - (2 - √3)/(-1 + √3) = - (1/2) (2 - √3)*(√3 + 1)   ...   (by rationalizing)
=> y = (1/2) (1 - √3)
Similarly, for x = - (2 + √3), y = (1/2)(1 + √3)
=> the points of intersection are
(-2 ± √3, (1 ∓ √3)/2).

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