GEC220 TUTORIALS: FUNCTIONS OF SEVERAL VARIABLES 1) In a coal processing plant the flow V of slurry along a pipe is given by: π£= πππ4 8Επ . If r and l both increase by 5% and P and Ε decrease by 10% and 30% respectively, using total differential, find the change in v and hence, approximate percentage change in V. Solutionπ£ = πππ4 8Επ ∴ π£ = π (π, π, π, Ε) Using total differential ππ£ ππ£ ππ£ πΕ ππ£ = ππ ππ + ππ ππ + ππ ππ + ππ πΕ ππ£ ππ ππ£ ππ ππ£ ππ ππ£ πΕ = 4ππ π3 8Επ πππ4 = − = 8Επ2 ππ4 8Επ πππ4 = − 8Ε2 π π (π) = 0.05 π π (π ) = 0.05 π π (π) = −0.1 π π (Ε) = −0.3 Ε Hence, the total differential equation becomes: ππ£ = 4ππ π3 8Επ 4 4 8Επ ∴ ππ£ = 4(0.05) πππ4 8Επ − 0.05 πππ4 8Επ = 0.35 8Επ πππ4 − 0.1 = (0.2 − 0.05 − 0.1 + 0.3) But π£ = 4 (0.05 π) − πππ2 (0.05 π ) + ππ (−0.1 π) − πππ2 (−0.3 Ε) 8Επ 8Ε π + 0.3 πππ4 8Επ ππ π4 8Επ πππ4 8Επ ∴ ππ£ = 0.35 π£ ii) Thus, % change in π£ = = 0.35 π£ π£ ππ£ π£ π₯ 100 π₯100 = 35 ∴ The % error in π£ is 35 % 1 πππ4 8Επ 2) The rate of flow of gas in a pipe, π£ is given by: π£ = πΆ π 1 ⁄2 π −5 ⁄6 where πΆ is a constant, π is the diameter of the pipe and π is the thermodynamic temperature of the gas. When determining the rate of flow experimentally, π is measured and subsequently found to be in error by +1.4% of π, and π has an error of –1.8% of π. (i) Using total differential, determine the error (change) in the rate of flow, π£, based on the measured values. Leave your answer in terms of π£. (ii) Hence, determine the percentage error in the rate of flow. Solution π£ = πΆ π 1 ⁄2 π −5 ⁄6 πΆ = πΆπππ π‘πππ‘ ∴ π£ = π (π, π) Using total differential ππ£ ππ£ ππ£ = ππ π(π ) + ππ ππ ππ£ ππ ππ£ 1 = πΆ (2 π −1⁄2 ) (π −5 ⁄6 ) 5 = πΆ (π 1 ⁄2 ) (− 6 π −11 ⁄6 ) π (π ) = 0.014 π π (π ) = −0.018 π Hence, the total differential equation becomes: ππ 1 5 ππ£ = πΆ (2 π −1 ⁄2 ) (π −5 ⁄6 ) (0.014 π ) + πΆ (π 1 ⁄2 ) (− 6 π −11 ⁄6 ) (−0.018 π) ∴ ππ£ = 0.007 πΆ (π 1 ⁄2 ) (π −5 ⁄6 ) + 0.015 πΆ (π 1 ⁄2 ) (π−5 ⁄6 ) = 0.022 πΆ (π 1 ⁄2 ) (π −5 ⁄6 ) But π£ = πΆ π 1 ⁄2 π −5 ⁄6 ∴ ππ£ = 0.022 π£ iii) Thus, % change in π£ = = 0.022 π£ π£ ππ£ π£ π₯ 100 100 = 2.2 ∴ The % error in π£ is 2.2 % 3) In a right-angled triangle, π denotes the hypotenuse, while π and π represent its other two sides. If π is increasing at the rate of 2 cm/s while π is decreasing at the rate of 3 cm/s. Using chain rule, calculate the rate at which π is changing when π = 5 cm and π = 3 cm. Hint: use Pythagoras theorem to relate π to π and π. SOLUTION From Pythagoras theorem, π 2 = π2 + π2 2 1 ∴ π = (π 2 − π 2 ) 2 π = π (π, π) (π, π ) = π (π‘) ππ ππ‘ =? Using chain rule ππ ππ = ππ‘ ππ . ππ ππ ππ‘ 1 = (π 2 − ππ 2 + π ππ ππ 1 2 )−2 . ππ ππ‘ β 2π 1 = π β (π 2 − π2 )−2 π = 2 2 √π −π ππ ππ 1 1 = (π 2 − π2 )−2 β (−2π) 2 1 = −π β (π 2 − π2 )−2 −π = 2 2 √π −π ππ = 2 ππ/π ππ‘ ππ ππ‘ = −3 ππ/π π = 5ππ π = 3 ππ ο ππ ππ‘ π = √π 2 −π2 . (2 ) + −π √π 2 −π2 . (−3) When π = 5ππ and π = 3 ππ ππ ππ‘ = 5 √5 2 −3 2 = = 10 4 19 4 . (2 ) + + −3 √5 2 −3 2 . (−3) 9 4 ππ = 4.75 ππ π is increasing at the rate of 4.75 ππ/π 4) Using chain rule, find the rate of change of the total surface area, π,of a right circular cone at the instant when thebase radius, π, is 5cmand the height, β, is 12cmif theradius is increasing at 0.5cm/s and the heightis decreasing at 1.5 cm/s. Given that π = π π2 + ππ √(π2 + β2 ). SOLUTION S= ππ2 + πr √(π2 + β2 ) =π π2 + ππ(π2 + β2 )1/2 π = π (π, β) 3 (π, β ) = π (π‘) ππ ππ‘ =? Using chain rule ππ ππ = ππ‘ ππ ππ ππ . ππ‘ + πβ . ππ‘ πβ 1 π ((π2 +β2 )2 ) ππ = 2ππ + ππ β ππ ππ 1 =2ππ + ππ. (π2 + β 2 ππ2 + (π2 +β2 )1/2 =2ππ ππ 1 + ( π 2 + β2 ) 2 . 1 2 ) −2 π ( ππ) ππ β 2π + (π2 + β2 )1/2. . π + π (π2 + β2 )1/2 1 = ππ. 2 ( π2 + β2 )−1/2 . 2β πβ 1 = ππβ β (π2 + β2 )−2 Given: ππ = 0.5 ππ/π ππ‘ πβ ππ‘ = −1.5 ππ/π π = 5ππ β = 12ππ 1 ο (π2 + β2 )2 = (52 + 122 )1/2 = √(25 + 144) = √169 = 13 ο ππ ππ ( )2 π 5 = 2π(5) + 13 + π (13) 25 =10 π + 13 π + 13π 130π+25π+169π = ππ = πβ ππ ο ππ‘ 13 π (( 5) ( 12) ) 13 = 324 13 = = 60 13 = 324π 13 π 60 π. (0.5) + 13 π. (−1.5) 162 13 72 π− 90 13 π =13 π = 17.4 ππ/π −1 5) Given that ππ + ππ − ππ + ππ − ππ = π using the general implicit formula, find ππ§ ππ₯ πππ ππ§ ππ¦ SOLUTION ππ + ππ − ππ + ππ − ππ − π = π 4 ππ§ = − ππ₯ ππ§ ππ¦ = − πΉπ₯ πΉπ§ πΉπ¦ πΉπ§ πΉπ₯ = 2π₯ + π§ πΉπ¦ = 2π¦ − π§ πΉπ§ = −2π§ + π₯ − π¦ ∴ ππ§ ππ₯ = − =( 2π₯+π§ −2π§+π₯ −π¦ 2π₯ +π§ 2π§−π₯ +π¦) ππ§ 2π¦ − π§ = − ππ¦ −2π§ + π₯ − π¦ =( 2π¦−π§ 2π§−π₯ +π¦) 6) Given that π π’ π ππ π = π¦, using the general implicit formula, find SOLUTION: π π’ π ππ π = π¦, ∴ π π’ π ππ π − π¦ = 0 Using general implicit formula, ππ’ ππ¦ ππ’ ππ£ ππ£ ππ¦ πΉπ¦ = − πΉπ’ πΉπ£ = − = − πΉπ’ πΉπ¦ πΉπ£ πΉπ’ = π π’ π ππ π πΉπ¦ = −1 πΉπ£ = π π’ πππ π ∴ ππ’ ππ¦ = − = (−1 ) ππ’ π ππ π 1 ππ’ π ππ π 5 ππ’ ππ¦ ππ’ , ππ£ πππ ππ£ ππ¦ ππ’ π π’ πππ π πππ π = − π’ =− ππ£ π π ππ π π ππ π =− ππ£ ππ¦ 1 π‘ππ π (−1 ) =− π π’ πππ π 1 = ππ’ π ππ π 7) Using general implicit formula, find ππ§ ππ₯ and ππ§ ππ¦ for the function: π₯ 3 π§ 2 − 5π₯π¦ 5 π§ = π₯ 2 + π¦ 3 3 2 5 2 π₯ 3 π§ 2 − 5π₯π¦ 5 π§ = π₯ 2 + π¦ 3 3 π₯ π§ − 5π₯π¦ π§ − π₯ − π¦ = 0 Using general implicit formula, ππ§ π = − ππ₯ ππ₯ π§ π ππ§ = − ππ¦ ππ¦ π§ ππ₯ = 3π₯ 2 π§ 2 − 5π¦ 5 π§ − 2π₯ ππ¦ = −25π₯π¦ 4 π§ − 3π¦ 2 ππ§ = 2π₯ 3 π§ − 5π₯π¦ 5 ∴ ππ§ ππ₯ ππ§ ππ¦ =− = − ππ₯ ππ§ ππ¦ ππ§ = − = (3π₯2 π§2 −5π¦5 π§−2π₯) 2π₯3 π§−5π₯π¦5 25π₯π¦4 π§+3π¦2 2π₯3 π§−5π₯π¦5 ( −πΌ ) 8) A possible equation of state for a gas takes the form: ππ = π π π ππ π in which πΌ and R ππ ππ are constants. Using the general implicit formula, determine expressions for , πππ ππ ππ ππ ππ SOLUTION ππ = π π π −πΌ ) ππ π ( 6 ∴ ππ − π π π −πΌ ) ππ π ( =0 Using the general implicit formula, ππ ππ ππ ππ ππ ππ = − πΉπ = − πΉπ = − πΉπ πΉπ πΉπ πΉπ π [ππ − π π π πΉπ = −πΌ ( ) ππ π ] ππ = π π [ππ − π π π πΉπ = ππ π[π = π − π π =π− πΌ π [( −πΌ ) π −1 ] π π ] ππ πΌ = π − π π [ π 2 π 2 π π ( πΉπ = π ( −πΌ ) ππ π ] −πΌ ) ππ π π [ππ − π π π = −πΌ ( ) ππ π ] −πΌ ( ) ππ π ] ππ π (ππ ) ππ ( −πΌ ) − π [−π π π ππ π ] ππ Using product rule to differentiate the R.H.S: [ ( −πΌ) π −1] ] π [ π ππ = 0 − [π π = − [π π β = −π β π ∴ ππ πΌ ππ π ( π 2 −πΌ ) ππ π [ −πΌ ) ππ π ( πΌ ππ π + π −πΌ ) π [ π π] ππ π ( +π βπ ππ ( −πΌ ) ππ π ] + 1] ππ· π = − π½ ππ½ ππ· ( −πΆ ) = − πΆ π·− π π π½πΉπ» π½ π½ 7 ] ππ πΉπ = − =− ππ πΉπ = −π β π −πΌ ( ) ππ π π− [ πΌ + 1] ππ π −πΌ πΌ (ππ π ) π 2 π ( −πΆ ) πΆ πΉβπ π½πΉπ» [ +π] π½πΉπ» −πΆ ) πΆ ( π·− π π π½πΉπ» π½ ππ πΉπ π = − = − −πΌ πΌ ( ) ππ πΉπ −π β π ππ π [ + 1] ππ π = π½ ( −πΆ ) πΆ πΉβπ π½πΉπ» [ +π] π½πΉπ» 9) An equation for heat generated π» is π» = π 2 π π‘. Using total differential, determine the percentage error in the calculated value of π» if the error in measuring current π is +2 %, the error in measuring resistance π is −3% and the error in measuring time t is +1%. 10) A rectangular box has sides of length x cm, y cm and z cm. Sides x and z are expanding at rates of 0.3 cm/s and 0.5 cm/s respectively and side y is contracting at a rate of 0.2 cm/s. Using chain rule, determine the rate of change of volume when x is 3 cm, y is 1.5 cm and z is 6 cm. 11) The power π dissipated in a resistor is given by π = πΈ2 π . If πΈ = 200 π£πππ‘π and π = 8 πβππ , using total differential, find the change in π resulting from a drop of 5 π£πππ‘π in πΈ and an increase of 0.2 πβππ in π . 12) The radius of a cylinder increases at the rate of 0.2 ππ/π while the height decreases at the rate of 0.5 ππ π . Using chain rule, find the rate at which the volume is changing at the instant when π = 8 ππ and β = 12 ππ. 8 13) The radius, π, of a cylindrical can is reduced by 20 % and its height, β, increased by 80 %. Using total differential, determine the change in volume when (π, β) = (1, 5). 14) The deflection π¦ at the centre of a circular plate suspended at the edge and uniformly loaded is given by π¦ = ππ€π4 π‘3 , where π€ = π‘ππ‘ππ ππππ, π = ππππππ‘ππ ππ ππππ‘π, π‘ = π‘βππππππ π πππ π ππ π ππππ π‘πππ‘. Calculate the approximate percentage change in π¦ if π€ is increased by 3 percent, π is decreased by 2 ½ and π‘ is increased by 4 πππππππ‘. 15) Given u2 – v2 -y = 0, using the general implicit formula, find 16) Given that (π + find ππ ππ , ππ ππ and π π2 π£2 ππ ππ ππ’ ππ£ , ππ’ ππ¦ ππ£ and ππ¦ ) (π£ − ππ) − ππ π = 0, using the general implicit formula, hence show that ππ ππ ππ x ππ x ππ ππ = -1 ππ’ ππ’ 17) Given eu cos V – x = 0, using the general implicit formula, find ππ₯ , ππ£ , πππ 9 ππ£ ππ₯ .