Word Problems - Derivatives

First, we can create an expanded polynomial:

\(y = x^3 + 3x^2\)

Next, we can differentiate this function by using Power Rule on both terms:

\(y' = \textcolor{red}{3}(x^{\textcolor{red}{3}-1}) + \textcolor{red}{2}(3x^{\textcolor{red}{2}-1})\)

\(y' = 3x^2 + 6x\)

After, we can set the derivative equal to the slope of the tangent. We can then shift all terms onto the same side to determine the \(x\)-intercepts of the tangent:

\(24 = 3x^2 + 6x\)

\(0 = 3x^2 + 6x - 24\)

\(0 = 3(x^2 + 2x - 8)\)

\(0 = 3(x - 2)(x + 4)\)

\(x = 2, -4\)

Finally, we can substitute these \(x\)-values into the original function to determine their respective \(y\)-coordinates:

\(y = x^3 + 3x^2\)

First, we can determine the \(y\)-coordinate for \(x = -4\):

\(y_1 = (\textcolor{red}{-4})^3 + 3(\textcolor{red}{-4})^2\)

\(y_1 = -64 + 3(16)\)

\(y_1 = -64 + 48\)

\(y_1 = -16\)

\(P_1 = (-4, -16)\)

Next, we can determine the \(y\)-coordinate for \(x = 2\):

\(y_2 = (\textcolor{red}{2})^3 + 3(\textcolor{red}{2})^2\)

\(y_2 = 8 + 3(4)\)

\(y_2 = 8 + 12\)

\(y_2 = 20\)

\(P_2 = (2, 20)\)

Therefore, we can determine that the 2 points on the graph where the tangent is \(24\) are \((-4, -16)\) and \((2, 20)\).

i. First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives:

\(\textcolor{red}{f(x) = 50000 + 6t}\)

\(\textcolor{red}{f'(x) = 6}\)

\(\textcolor{blue}{g(x) = 1 + 0.4t}\)

\(\textcolor{blue}{g'(x) = 0.4}\)

Next, we must use Quotient Rule in order to determine the respective Rates of Change:

\(V'(t) = \cfrac{\textcolor{blue}{g(t)}\textcolor{red}{f'(t)} - \textcolor{red}{f(t)}\textcolor{blue}{g'(t)}}{\textcolor{blue}{g(t)}^2}\)

\(V'(t) = \cfrac{(\textcolor{blue}{1 + 0.4t})(\textcolor{red}{6}) - (\textcolor{red}{50000 + 6t})(\textcolor{blue}{0.4})}{(\textcolor{blue}{1 + 0.4t})^2}\)

\(V'(t) = \cfrac{6 + 2.4t - (20000 + 2.4t)}{(1 + 0.4t)^2}\)

Then, we can simplify the expression by expanding and collecting like terms:

\(V'(t) = \cfrac{6 + 2.4t -(20000 + 2.4t)}{(1 + 0.4t)^2}\)

\(V'(t) = \cfrac{6 + 2.4t - 20000 - 2.4t}{(1 + 0.4t)^2}\)

\(V'(t) = \cfrac{-19994}{(1 + 0.4t)^2}\)

Now, we can substitute the respective year values for \(t\) to determine the corresponding rate of change values.

First, we can determine the rate of change of the car's value at \(2\) years:

\(V'(\textcolor{red}{2}) = \cfrac{-19994}{(1 + 0.4(\textcolor{red}{2}))^2}\)

\(V'(\textcolor{red}{2}) = \cfrac{-19994}{(1.8)^2}\)

\(V'(\textcolor{red}{2}) = \cfrac{-19994}{3.24}\)

\(\textcolor{red}{V'(2) = -$6170.99/\text{year}}\)

Next, we can determine the rate of change of the car's value at \(5\) years:

\(V'(\textcolor{green}{5}) = \cfrac{-19994}{(1 + 0.4(\textcolor{green}{5}))^2}\)

\(V'(\textcolor{green}{5}) = \cfrac{-19994}{(3)^2}\)

\(V'(\textcolor{green}{5}) = \cfrac{-19994}{9}\)

\(\textcolor{green}{V'(5) = -$2221.56/\text{year}}\)

Finally, we can determine the rate of change of the car's value at \(5\) years:

\(V'(\textcolor{blue}{7}) = \cfrac{-19994}{(1 + 0.4(\textcolor{blue}{7}))^2}\)

\(V'(\textcolor{blue}{7}) = \cfrac{-19994}{(3.8)^2}\)

\(V'(\textcolor{blue}{7}) = \cfrac{-19994}{14.44}\)

\(\textcolor{blue}{V'(7) = -$1384.63/\text{year}}\)

Therefore, we can determine that the respective rates of change of the car's value at \(2\), \(5\) and \(7\) years are \(-$6170.99/\text{year}\), \(-$2221.56/\text{year}\) and \(-$1384.63/\text{year}\).

---

ii. In order to determine the initial value of the car we can substitute \(0\) for \(t\) in the original expression:

\(V(t) = \cfrac{50000 + 6t}{1 + 0.4t}\)

\(V(\textcolor{teal}{0}) = \cfrac{50000 + 6(\textcolor{teal}{0})}{1 + 0.4(0)}\)

\(V(\textcolor{teal}{0}) = \cfrac{50000}{1}\)

\(V(\textcolor{teal}{0}) = $50000\)

Therefore, we can determine that the initial value of the car was \($50000\).

---

iii. The values in Q1 show that the car depreciates in value significantly over time. Therefore, it would make sense financially to buy a used one rather than a new one.

First, we can write this problem as a Chain Rule expression:

\(\cfrac{dC}{dt} = \cfrac{dC}{dq}\cdot \cfrac{dq}{dt}\)

Next, we can rewrite \(C(q)\):

\(C(q) = 300q + 0.2q^{\frac{1}{2}} + 20q^{-1}\)

Now, we can differentiate \(C(q)\) by using Implicit Differentiation:

\(C'(q) = \cfrac{dC}{dt}(300q)\cdot \cfrac{dq}{dt} + \cfrac{dC}{dt}(0.2q^{\frac{1}{2}})\cdot \cfrac{dq}{dt} + \cfrac{dC}{dt}(20q^{-1})\cdot \cfrac{dq}{dt}\)

\(C'(q) = 300\cdot \cfrac{dq}{dt} + 0.1q^{-\frac{1}{2}}\cdot \cfrac{dq}{dt} -20q^{-2}\cdot \cfrac{dq}{dt}\)

\(C'(q) = 300\cdot \cfrac{dq}{dt} + \cfrac{0.1}{q^{\frac{1}{2}}}\cdot \cfrac{dq}{dt} \cfrac{-20}{q^2}\cdot \cfrac{dq}{dt}\)

Next, we can substitute \(4\) for \(t\) in \(q(t)\) to determine the # of units produced at that time:

\(q(4) = (4)^3 - \cfrac{2}{\sqrt{4}}\)

\(q(4) = 64 - 1\)

\(q(4) = 63\)

After, we can differentiate \(q(t)\). We can start by rewriting the equation as such:

\(q(t) = t^3 - 2t^{-\frac{1}{2}}\)

Now, we can differentiate \(q(t) = t^3 -\cfrac{2}{t}\) by using Power Rule:

\(q'(t) = 3t^2 + t^{-\frac{3}{2}}\)

\(q'(t) = 3t^2 + \cfrac{1}{t^{\frac{3}{2}}}\)

We can substitute \(4\) (the elapsed time in hours) for \(t\) to determine the # of units/hour within that time:

\(q'(4) = 3(4)^2 + \cfrac{1}{(\sqrt{4})^3}\)

\(q'(4) = 48 + \cfrac{1}{8}\)

\(q'(4) = \cfrac{385}{8}\)

Finally, we can substitute \(\cfrac{385}{8}\) in for \(\cfrac{dq}{dt}\) and \(64\) for \(q\) in the Cost equation to determine the rate the cost is changing:

\(C'(q) = 300 \left(\cfrac{385}{8}\right) + \cfrac{0.1}{(63)^{\frac{1}{2}}} \left(\cfrac{385}{8}\right) - \cfrac{20}{(63)^2}\left(\cfrac{385}{8}\right)\)

\(C'(q) = 14437.5 + 0.61 - \cfrac{20}{3969}(\cfrac{385}{8})\)

\(C'(q) = 14437.87\)

Therefore, we can determine that the rate of the cost is increasing \($14437.87/[\text{hr}]\) \(4\) hours after production commences.