Differentiation: Bases other than e, separation of variables, slope fields

## Bases Other Than $e$: Differentiation & Integration ### General Exponential and Logarithmic Functions While $e^x$ and $\ln x$ are the "natural" bases in calculus due to their simple derivatives, real-world data often uses other bases (e.g., base 2 for computing, base 10 for pH or decibels). Any exponential $a^x$ can be rewritten as $e^{x \ln a}$, which dictates the unique scaling factors ($\ln a$) seen in their calculus rules. ### Calculus Rules for Base $a$ * **Derivatives of Exponential Functions:** $$\frac{d}{dx}[a^u] = (\ln a) a^u \frac{du}{dx}$$ * *Nuance:* The $\ln a$ is a constant multiplier. For $e^x$, this multiplier is $\ln e = 1$, which is why $e^x$ is unique. * **Integrals of Exponential Functions:** $$\int a^u du = \frac{1}{\ln a} a^u + C$$ * **Derivatives of Logarithmic Functions:** $$\frac{d}{dx}[\log_a u] = \frac{1}{u \ln a} \frac{du}{dx}$$ * *Context:* This is derived from the Change of Base Formula: $\log_a u = \frac{\ln u}{\ln a}$. * **Common Pitfalls:** * **Confusing Power Rule with Exponential Rule:** Do not use the power rule on $a^x$. $\frac{d}{dx}[5^x] \neq x \cdot 5^{x-1}$. * **Forgetting the Constant:** In differentiation, you **multiply** by $\ln a$; in integration, you **divide** by $\ln a$. ### Worked Examples: Base $a$ **Problem 1: Complex Base Differentiation** Find the derivative of $y = 5^{\sqrt{x}}$. * Identify $a = 5$ and $u = x^{1/2}$ $$y' = (\ln 5) \cdot 5^{\sqrt{x}} \cdot \frac{d}{dx}[x^{1/2}]$$ * Apply the general rule $$y' = (\ln 5) \cdot 5^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$$ * Final result $$y' = \frac{(\ln 5) 5^{\sqrt{x}}}{2\sqrt{x}}$$ **Problem 2: Integration of Base $a$** Evaluate $\int 7^x dx$. * Identify $a = 7$ $$\int 7^x dx = \frac{1}{\ln 7} 7^x + C$$ --- ## Differential Equations: Growth and Decay ### The Law of Exponential Change Many natural phenomena follow the rule that the rate of change of a quantity is directly proportional to the amount of the quantity present. This leads to the fundamental differential equation $\frac{dy}{dt} = ky$. ### Modeling Components & Applications * **The General Solution:** $y = Ce^{kt}$ * $C$: Initial amount (when $t=0$). * $k$: The relative growth rate. * **Half-Life ($t_{1/2}$):** The time required for a quantity to decay to half its initial size. $$k = \frac{\ln(0.5)}{t_{1/2}} \approx \frac{-0.693}{t_{1/2}}$$ * **Newton’s Law of Cooling:** The rate of change of an object's temperature $T$ is proportional to the difference between $T$ and the ambient temperature $K$. $$\frac{dT}{dt} = k(T - K) \implies T = K + (T_0 - K)e^{kt}$$ * **Scannable Tip:** If the problem says "Rate is proportional to the amount," immediately jump to $y = Ce^{kt}$. If it says "Rate is proportional to the difference," use Newton's model. ### Worked Example: Newton's Law of Cooling **Problem:** A cup of coffee at $180^\circ F$ is placed in a $70^\circ F$ room. After 2 minutes, it is $165^\circ F$. Find the temperature after 10 minutes. * Identify constants: $K = 70$, $T_0 = 180$. $$T(t) = 70 + (180 - 70)e^{kt} = 70 + 110e^{kt}$$ * Use $T(2) = 165$ to find $k$: $$165 = 70 + 110e^{2k} \implies 95 = 110e^{2k}$$ $$\frac{95}{110} = e^{2k} \implies k = \frac{\ln(95/110)}{2} \approx -0.0733$$ * Solve for $T(10)$: $$T(10) = 70 + 110e^{-0.0733(10)} \approx 70 + 110(0.480) \approx 122.8^\circ F$$ --- ## Differential Equations: Separation of Variables ### The Separable Process Separation of variables is a technique for solving first-order differential equations where the expression for $dy/dx$ can be factored into a function of $x$ and a function of $y$. ### Methodology and Execution 1. **Separate:** Rewrite the equation so all $y$ terms (including $dy$) are on one side and all $x$ terms (including $dx$) are on the other. $$\frac{1}{g(y)} dy = f(x) dx$$ 2. **Integrate:** Integrate both sides independently. 3. **Solve for $C$:** Use the initial condition (e.g., $y(0) = 4$) to find the specific value of the constant of integration. 4. **Isolate $y$:** Solve the resulting algebraic equation to get $y$ as an explicit function of $x$ if possible. * **Subtle Nuance:** The constant $C$ often becomes part of a coefficient after exponentiation. If $\ln|y| = x + C$, then $|y| = e^{x+C} = e^C e^x$, which simplifies to $y = Ae^x$. ### Worked Example: Separable Equation **Problem:** Solve $\frac{dy}{dx} = \frac{3x^2}{e^{2y}}$ with initial condition $f(0) = 0.5$. * Separate the variables: $$e^{2y} dy = 3x^2 dx$$ * Integrate both sides: $$\int e^{2y} dy = \int 3x^2 dx \implies \frac{1}{2}e^{2y} = x^3 + C$$ * Use initial condition $(0, 0.5)$: $$\frac{1}{2}e^{2(0.5)} = 0^3 + C \implies \frac{1}{2}e^1 = C \implies C = \frac{e}{2}$$ * Substitute $C$ back and solve for $y$: $$\frac{1}{2}e^{2y} = x^3 + \frac{e}{2} \implies e^{2y} = 2x^3 + e$$ $$2y = \ln(2x^3 + e) \implies y = \frac{1}{2}\ln(2x^3 + e)$$ --- ## Slope Fields ### Visualizing Differential Equations A slope field (or direction field) is a graphical tool used to visualize the solutions of a first-order differential equation $y' = f(x, y)$ without solving the equation analytically. ### Interpretation and Construction * **Components:** At various points $(x, y)$ in the plane, a short line segment is drawn with a slope equal to $f(x, y)$. * **Solution Curves:** To sketch a particular solution, start at a given point and "follow the flow" of the segments, keeping the curve tangent to the segments at every point. * **Analyzing Equilibrium:** * **Horizontal Segments:** Occur where $y' = 0$. These often represent equilibrium solutions or local extrema. * **Vertical Segments:** Occur where the derivative is undefined (asymptotes). * **Pattern Recognition for Matching:** * If segments are identical along vertical lines, the DE depends only on $x$ (e.g., $y' = x^2$). * If segments are identical along horizontal lines, the DE depends only on $y$ (e.g., $y' = y - 2$). * If segments show symmetry across the origin, look for functions like $y' = -x/y$ (circles). graph[-x/y][-4][4] *(Visual representation of a slope field for $y' = -x/y$, suggesting circular solution paths.)* ### Logic for Matching Examples **Scenario: Matching $y' = x + y$ to a field.** * Check the origin $(0,0)$: Slope should be $0+0=0$ (horizontal segment). * Check $(1, -1)$: Slope should be $1 + (-1) = 0$ (horizontal segment along the line $y = -x$). * Check the first quadrant: As $x$ and $y$ increase, the slopes should become increasingly positive (steeper). * Check the third quadrant: As $x$ and $y$ become more negative, the slopes should become increasingly negative.

Bases Other Than : Differentiation & IntegrationGeneral Exponential and Logarithmic Functions While and are the "natural" bases in calculus due to their simple derivatives, real-world data often uses other bases (e.g., base 2 for computing, base 10 for pH or decibels). Any exponential can be rewritten as , which dictates the unique scaling factors () seen in their calculus rules.Calculus Rules for Base Derivatives of Exponential Functions:Nuance: The is a constant multiplier. For , this multiplier is , which is why is unique.Integrals of Exponential Functions:Derivatives of Logarithmic Functions:Context: This is derived from the Change of Base Formula: .Common Pitfalls:Confusing Power Rule with Exponential Rule: Do not use the power rule on . .Forgetting the Constant: In differentiation, you multiply by ; in integration, you divide by .Worked Examples: Base Problem 1: Complex Base Differentiation Find the derivative of .Identify and Apply the general ruleFinal resultProblem 2: Integration of Base Evaluate .Identify Differential Equations: Growth and DecayThe Law of Exponential ChangeMany natural phenomena follow the rule that the rate of change of a quantity is directly proportional to the amount of the quantity present. This leads to the fundamental differential equation .Modeling Components & ApplicationsThe General Solution: : Initial amount (when ).: The relative growth rate.Half-Life (): The time required for a quantity to decay to half its initial size.Newton’s Law of Cooling: The rate of change of an object's temperature is proportional to the difference between and the ambient temperature .Scannable Tip: If the problem says "Rate is proportional to the amount," immediately jump to . If it says "Rate is proportional to the difference," use Newton's model.Worked Example: Newton's Law of CoolingProblem: A cup of coffee at is placed in a room. After 2 minutes, it is . Find the temperature after 10 minutes.Identify constants: , .Use to find :Solve for :Differential Equations: Separation of VariablesThe Separable ProcessSeparation of variables is a technique for solving first-order differential equations where the expression for can be factored into a function of and a function of .Methodology and ExecutionSeparate: Rewrite the equation so all terms (including ) are on one side and all terms (including ) are on the other. $$\frac{1}{g(y)} dy = f(x) dx$$Integrate: Integrate both sides independently.Solve for : Use the initial condition (e.g., ) to find the specific value of the constant of integration.Isolate : Solve the resulting algebraic equation to get as an explicit function of if possible.Subtle Nuance: The constant often becomes part of a coefficient after exponentiation. If , then , which simplifies to .Worked Example: Separable EquationProblem: Solve with initial condition .Separate the variables:Integrate both sides:Use initial condition :Substitute back and solve for :Slope FieldsVisualizing Differential EquationsA slope field (or direction field) is a graphical tool used to visualize the solutions of a first-order differential equation without solving the equation analytically.Interpretation and ConstructionComponents: At various points in the plane, a short line segment is drawn with a slope equal to .Solution Curves: To sketch a particular solution, start at a given point and "follow the flow" of the segments, keeping the curve tangent to the segments at every point.Analyzing Equilibrium:Horizontal Segments: Occur where . These often represent equilibrium solutions or local extrema.Vertical Segments: Occur where the derivative is undefined (asymptotes).Pattern Recognition for Matching:If segments are identical along vertical lines, the DE depends only on (e.g., ).If segments are identical along horizontal lines, the DE depends only on (e.g., ).If segments show symmetry across the origin, look for functions like (circles). (Visual representation of a slope field for , suggesting circular solution paths.)Logic for Matching ExamplesScenario: Matching to a field.Check the origin : Slope should be (horizontal segment).Check : Slope should be (horizontal segment along the line ).Check the first quadrant: As and increase, the slopes should become increasingly positive (steeper).Check the third quadrant: As and become more negative, the slopes should become increasingly negative.