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Magazine Chapter 4: Problem 22
Write the first and second derivatives of the function and use the second derivative to determine inputs at which inflection points might exist. \(g(t)=\frac{79}{1+36 e^{0.2 t}}+13\)
Short Answer
Expert verified
The potential inflection points are found by solving \( g''(t) = 0 \).
Step by step solution
01
Find the First Derivative
To find the first derivative of the function \( g(t) = \frac{79}{1+ 36 e^{0.2 t}} + 13 \), apply the quotient rule. Let \( u = 79 \) and \( v = 1 + 36 e^{0.2 t} \). The derivative of \( u \) is \( u' = 0 \), and the derivative of \( v \) is \( v' = 36 \times 0.2 e^{0.2 t} = 7.2 e^{0.2 t} \). The quotient rule is \( (\frac{u}{v})' = \frac{v u' - u v'}{v^2} \). So, \( g'(t) \) becomes:\[ g'(t) = \frac{0 \cdot (1 + 36 e^{0.2 t}) - 79 \cdot 7.2 e^{0.2 t}}{(1 + 36 e^{0.2 t})^2} = \frac{-568.8 e^{0.2 t}}{(1 + 36 e^{0.2 t})^2}. \]
02
Find the Second Derivative
Now, differentiate \( g'(t) = \frac{-568.8 e^{0.2 t}}{(1 + 36 e^{0.2 t})^2} \) using both the product and quotient rules. Since this involves a negative coefficient, treat it separately for clarity:Let \( a = -568.8 e^{0.2 t} \), and its derivative, using the chain rule, is \( a' = -568.8 \times 0.2 e^{0.2 t} = -113.76 e^{0.2 t} \). Now, let \( b = (1 + 36 e^{0.2 t})^2 \) and its derivative using the chain rule and product rule is \( b' = 2(1 + 36 e^{0.2 t}) \times 7.2 e^{0.2 t} \).\ Hence, \[ g''(t) = \frac{b \cdot a' - a \cdot b'}{b^2}. \] Substitute \( a' \), \( a \), \( b' \) and \( b \) to find \( g''(t) \): \[ g''(t) = \frac{ (1 + 36 e^{0.2 t})^2 \times (-113.76 e^{0.2 t}) + 568.8 e^{0.2 t} \times 2(1 + 36 e^{0.2 t}) \times 7.2 e^{0.2 t} }{((1 + 36 e^{0.2 t})^2)^2}. \] Evaluate to simplify further.
03
Determine Inflection Points
Inflection points occur where the second derivative \( g''(t) \) changes sign. Thus, solve \( g''(t) = 0 \) using the expression from Step 2. Simplify and set the numerator equal to zero. Solve the resulting equation for \( t \) to identify potential inflection points:\[ -113.76 e^{0.2 t} (1 + 36 e^{0.2 t})^2 + 568.8 e^{0.4 t} \times 2(1 + 36 e^{0.2 t}) = 0. \] Simplifying and solving this complex equation analytically or using numerical methods will give potential values of \( t \) for inflection points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
When we talk about derivatives in calculus, we are referring to the rate of change of a function. Specifically, the derivative tells us how much the function's value changes as we change the input slightly. In practical terms, it's like asking, "How fast is this value changing?" For the function given in the exercise and solution, the first derivative, denoted as \( g'(t) \), helps us understand how \( g(t) \) changes with respect to \( t \).
In our case, \( g'(t) \) involves the quotient rule because the function is a fraction. By finding the first derivative, you open the door to more advanced insights, like determining the slope of the tangent line to the curve at any point, or assessing whether the function is increasing or decreasing.
In our case, \( g'(t) \) involves the quotient rule because the function is a fraction. By finding the first derivative, you open the door to more advanced insights, like determining the slope of the tangent line to the curve at any point, or assessing whether the function is increasing or decreasing.
Inflection Points
Inflection points are considered a crucial aspect of the analysis of curves in calculus. They are points on the graph of a function where the curvature changes direction. Think of it as the transition from a "smile" to a "frown" shape on a curve, or vice versa. To find inflection points, you typically need to look at the second derivative of the function, as it reveals information about concavity鈥攚hether the graph of the function is opening upwards or downwards.
An inflection point occurs when a second derivative changes sign. For the function in our exercise, finding \( g''(t) \) and determining where it equals zero or changes sign would identify potential points of inflection. These points provide valuable insights because they highlight where the function's growth rate is changing in the most meaningful way.
An inflection point occurs when a second derivative changes sign. For the function in our exercise, finding \( g''(t) \) and determining where it equals zero or changes sign would identify potential points of inflection. These points provide valuable insights because they highlight where the function's growth rate is changing in the most meaningful way.
Quotient Rule
The quotient rule is a method used in calculus for finding the derivative of a quotient of two functions. It may seem daunting at first but is straightforward once you get the hang of it. If you have a function \( h(t) = \frac{u(t)}{v(t)} \), the quotient rule states that the derivative \( h'(t) \) is given by:
In our case, for \( g(t) \), the quotient rule helps us compute \( g'(t) \) accurately by considering both the top and bottom parts of the fraction. The application of the rule reveals how sophisticated derivatives in calculus can become, especially when applied to complex expressions like quotients.
- Find the derivative of the numerator \( u' \)
- Find the derivative of the denominator \( v' \)
- Plug into the formula: \( h'(t) = \frac{v(t) u'(t) - u(t) v'(t)}{(v(t))^2} \)
In our case, for \( g(t) \), the quotient rule helps us compute \( g'(t) \) accurately by considering both the top and bottom parts of the fraction. The application of the rule reveals how sophisticated derivatives in calculus can become, especially when applied to complex expressions like quotients.
Calculus
Calculus is a branch of mathematics focused on studying change through derivatives and integration. It's about understanding how things evolve over time or space. You can think of it as a toolkit for analysis and problem-solving involving dynamic systems.
Derivatives, and consequently topics like the second derivative and inflection points, are one half of calculus, often referred to as differential calculus. The other half, integral calculus, deals with sums and areas. In our function exercise, through calculus, you learn not just to manipulate formulas but to visualize and predict how a given system behaves through functions like \( g(t) \).
Derivatives, and consequently topics like the second derivative and inflection points, are one half of calculus, often referred to as differential calculus. The other half, integral calculus, deals with sums and areas. In our function exercise, through calculus, you learn not just to manipulate formulas but to visualize and predict how a given system behaves through functions like \( g(t) \).
- Calculus allows you to predict future trends.
- It provides tools to optimize and determine maxima and minima.
- Understanding these concepts helps in fields ranging from engineering to economics.
