for some tasks, a sigmoid learning curve is used to model proficiency: 4) 74%® F
ID: 2849385 • Letter: F
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for some tasks, a sigmoid learning curve is used to model proficiency: 4) 74%® Fri 11:42 AM a M22- Derivatives and Curve Sketching Updates Available Do you want to restart to install these updates now or try tonight? /2439059?module-item-id=3331070 Restart Later Grades Calendar Resources canvas 401-F15> Assignments M22- Derivatives and Curve Sketching atives & Curve Sketching Show Intro/Instructions For some tasks, a sigmoid leaming curve is used to model proficieney. For some tasks, a sigmoid learning curve is used to model proficiency. Suppose the percent proficiency is given by the 100 1 + 50e-04t , where t is days of practice. 1 + 50e-04 model P(t) = Find the inflection point, the point where additional practice starts producing diminishing returns. Give your answer accurate to at least 2 decimal places. Preview days License Points possible: 1 This is attempt 1 of 3. Submit 1]Explanation / Answer
p(t) = 100/(1 +50e-0.4t)
p '(t) = [ (0)(1 +50e-0.4t) -(100)50(e-0.4t) (-0.4)]/(1 +50e-0.4t)2 (u/v)' = [u'v + uv']/v2 ; d/dx f(g(x)) = f '(g(x)) g '(x)
==> p '(t) = 2000(e-0.4t)/(1 +50e-0.4t)2
==> p ''(t) = [2000e-0.4t(-0.4)(1 +50e-0.4t)2 - 2000(e-0.4t)(2)(1 +50e-0.4t)(50e-0.4t)(-0.4)]/(1 +50e-0.4t)4
==> p ''(t) = [-800e-0.4t(1 +50e-0.4t) + 1600(e-0.4t)(50e-0.4t)]/(1 +50e-0.4t)3
==> p ''(t) = [-800e-0.4t -40000e-0.8t + 80000e-0.8t]/(1 +50e-0.4t)3
==> p ''(t) = [ 40000e-0.8t - 800e-0.4t]/(1 +50e-0.4t)3
inflection point ==> p ''(t) = 0
==> 40000e-0.8t - 800e-0.4t = 0
==> 50e-0.8t = e-0.4t
==>50 = e0.4t
==> e0.4t = 50
Apply natural logarithms on both sides
==> ln e0.4t = ln 50
==> 0.4t = ln50
==> t = (ln50)/0.4
==> t = 9.78 days
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