Can you help me solve this? **Differentiating a Weight Function**Consider the weight function:\[ W(t) = 26.7(1 - e^{-0.18t})^3, \]where \( W \) is in kilograms and \( t \) is in years.**Step a: Differentiate the Weight Function**The first derivative of the weight function is:\[ W'(t) = \frac{7209 \cdot e^{-(27t)/50} \cdot (e^{(9t)/50} - 1)^2}{500} \]**Find the Second Derivative**The second derivative of the weight function is:\[ W''(t) = \frac{64881 \cdot e^{-(9t)/25} \cdot (e^{(9t)/50} - 1)}{12500} - \frac{194643 \cdot e^{-(27t)/50} \cdot (e^{(9t)/50} - 1)^2}{62500} \]**Points of Inflection**Determine the \( t \) and \( W \) values for any points of inflection \( (t > 0) \).- \( t_i = \) [Fill in the value] yr.- \( W(t_i) = \) [Fill in the value] kg.**Maximum Rate of Weight Gain**The trout is gaining weight most rapidly at the point of inflection. Find this most rapid rate of growth.- \( W'(t_i) = \) [Fill in the value] kg/yr.

The given function is W(t)=26.71-e-0.18t3. Now, differentiate the function as follows: W '(t)=26.7·3·1-e-0.18t2·ddt1-e-0.18 [Apply power rule and chain rule] ⇒W '(t)=80.11-e-0.18t2·0.18·e-0.18t [Applying power rule again for exponential function] ⇒W ' (t)=14.418·e-0.18t·1-e-0.18t2 Differentiate again for the second derivative: W ''(t)=14.418·e-0.18t·ddt1-e-0.18t2+14.418·1-e-0.18t2·ddte-0.18t[Applying product rule] ⇒W ''(t)=5.19048e-0.36t(1-e-0.18t)-2.59524e-0.18t(1-e-0.18t)2 For the inflection point the second derivative is either not defined or zero. Here, solve W ''(t)=0. ⇒5.19048e-0.36t1-e-0.18t-2.59524e-0.18t1-e-0.18t2=0 ⇒1-e-0.18t5.19048e-0.36t-2.59524e-0.18t1-e-0.18t=0 ⇒1-e-0.18t7.78572 e-0.36t-2.59524e-0.18t=0 When 1-e-0.18t=0 ⇒e-0.18t=1 ⇒t=0 When 7.78572 e-0.36t-2.59524e-0.18t=0 ⇒7.78572e-0.36t=2.59524e-0.18t ⇒e-0.18t=2.595247.78572 ⇒-0.18t=ln2.595247.78572 ⇒t=-10.18·ln2.595247.78572 ⇒t=6.1034 yr. Now, W(t)=26.71-2.595247.785723=7.91111 kg. The most rapid rate of growth is W '(t)=14.4182.595247.785721-2.595247.785722=2.136 kg/yr.

Math

Calculus

W (t) = 26.7(1 – e 0.18t )3, where W is in kg and t is in years. a. Differentiate this weight function. W'(t) = (7209*e^(-(27*t)/50)*(e^((9*t)/50)-1)^2)/500 Find its second derivative. W" (t) = (64881*e^(-(9*t)/25)*(e^((9*t)/50)-1))/12500-(194643*e^(-(27*t)/50)*(e' Give both the t and W values for any points of inflection (t > 0). ti yr. W (t;) = kg. The trout is gaining weight most rapidly at this point of inflection. Find this most rapid rate of growth. W (t;) = kg/yr.

W (t) = 26.7(1 – e 0.18t )3, where W is in kg and t is in years. a. Differentiate this weight function. W'(t) = (7209e^(-(27t)/50)(e^((9t)/50)-1)^2)/500 Find its second derivative. W" (t) = (64881e^(-(9t)/25)(e^((9t)/50)-1))/12500-(194643e^(-(27t)/50)*(e' Give both the t and W values for any points of inflection (t > 0). ti yr. W (t;) = kg. The trout is gaining weight most rapidly at this point of inflection. Find this most rapid rate of growth. W (t;) = kg/yr.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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