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# Assignment on Differential Calculus

Added on - 21 Feb 2021

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Differential Calculus
TABLE OF CONTENTTASK 1...........................................................................................................................................1A. Differentiate the algebraic function.......................................................................................1B. Differentiating using quotient rule.........................................................................................1C. Differentiating logarithmic function......................................................................................2D. Differentiating using product rule..........................................................................................2TASK 2...........................................................................................................................................3B. Area enclosed by curves.........................................................................................................3TASK 3............................................................................................................................................4A. Calculating Ɵ1.......................................................................................................................4B.) Calculating temperature of cooling object............................................................................5TASK 4............................................................................................................................................6A Least area of lidless box..........................................................................................................6B Height and radius of cylinder with least surface area.............................................................7
TASK 1A. Differentiate the algebraic functionEquation linking distance and time is given by:x = 3t + 0.5 at²a = accelerationt = timev = velocitya = 1.6 m/s²t = 5 secondsSolutionVelocity (v) = Rate of change of distance = dx/dtv = d (3t + 0.5 at² ) / dtv = 3 + 0.5a *2t(a is constant = 1.6, as given)v = 3 + 1.6tat t = 5 secondsvelocity = 3 +(1.6*5) = 11 m/secondB. Differentiating using quotient ruley= [e^(2t) ] / sin 5tSolutionAs per quotient rule:If u and v are function of t thendu / dv = (du *v ) - (u*dv) / v² where du = du/dt and dv = dv/dtIn the given situationu = e^(2t)v = sin 5tdu / dt = 2e^(2t)dv / dt = 5cos 5tOn substituting these values we getdy = [2e^(2t) * sin 5t ] - [e^(2t) * 5cos 5t ] / (sin 5t)²dy = e^(2t) [2sin 5t - 5cos 5t] / (sin 5t)²1
C. Differentiating logarithmic functiony = ln [(x-2) (x+1) / (x-1) (x+3) ]SolutionUsing the formula: d (ln x) / dx = 1/xUsing the property of logarithmic functions we have:ln (a/b) ln a – ln bln (ab) = ln a * ln bSimilarly for function y:ln (x-2) (x+1) / (x-1) (x+3) = ln [(x-2) (x+1) ] - ln [(x-1) (x+3)]It can be further simplified as =ln (x-2) + ln (x+1) - [ln (x-1) + ln (x+3)]y= ln (x-2) + ln (x+1) - ln (x-1) - ln (x+3)Using differentiation property we have:dy /dx = [1 / (x-2)] + [1 / (x+1)] - [1 / (x-1)] - [1 / (x+3) ]D. Differentiating using product rulev = 4t sin 2twhen t = 0.5 secondsSolutionAccording to product rule if a and b are two functions of x thend (ab) /dx = a'b + ab' wherea' = da /dxand b = db /dxv = 4t sin 2tIn the given function of voltage a = t and b = sin 2tOn applying differentiation property of products we can write:dv / dt = 4 * [ t * d (sin 2t ) + d (t) * sin 2t ] dtdv / dt = 4 * [ t * 2 * cos 2t ] + [sin 2t ]dv / dt = 4 * [2t cos 2t + sin 2t ]dv / dt = 4 * [2t cos 2t + sin 2t ]At t = 0.5 secondsRate of voltage change across capacitor = 4 [sin 2(0.5) + 2(0.5) cost 2(0.5)]2