Problem 12 - Applied Calculus

Question

Find the point where the following curve is steepest: $$y=\frac{50}{1+6 e^{-2 t}} \quad \text { for } t \geq 0$$

Step-by-Step Solution

Answer

The curve is steepest at \( t = 0 \) where the derivative is maximum.

Step 1: Understand the problem

To find where the curve is steepest, you need to determine the point where the slope of the curve is at its maximum. This involves finding the maximum value of the derivative.

Step 2: Differentiate the function

The given function is \( y = \frac{50}{1 + 6 e^{-2t}} \). Let's find the derivative \( \frac{dy}{dt} \). Use the quotient rule: \( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = 50 \) and \( v = 1 + 6e^{-2t} \).

Step 3: Apply the quotient rule

Using the quotient rule, find the derivative: \( \frac{dy}{dt} = \frac{0(1 + 6e^{-2t}) - 50(-12e^{-2t})}{(1 + 6e^{-2t})^2} = \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \).

Step 4: Simplify the derivative

The derivative simplifies to \( \frac{dy}{dt} = \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \). To find where the curve is steepest, we need to find the maximum value of this derivative.

Step 5: Find the critical points

To find maximum points, consider setting the derivative of \( \frac{dy}{dt} \) equal to zero and solve for \( t \), but since \( 600e^{-2t} \) is always positive, solve \( \frac{d}{dt}\left( \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \right) = 0 \).

Step 6: Differentiate derivative to find maximum

Compute the derivative of \( \frac{dy}{dt} \) to find \( \frac{d^2y}{dt^2} \). This involves applying the quotient rule again to the expression \( \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \). Set this equal to zero to find the points that might be a maximum or minimum.

Step 7: Solve for the critical points

After differentiating, solve \( \frac{d^2y}{dt^2} = 0 \), which generally involves finding \( t \) by setting the numerator of the second derivative to zero (as the denominator can't be zero for \( t \geq 0 \)).

Step 8: Evaluate critical points

After finding potential \( t \) values, plug them back into the original function to determine the corresponding \( y \) values and ensure that the second derivative test confirms it is a maximum.

Step 9: Confirm the steepest point

From the calculations and the second derivative test, verify which \( t \) gives the maximum slope (if multiple solutions for \( t \) ensure it is a maximum by testing the second derivative).

Key Concepts

Quotient Rule in Calculus

The quotient rule is a technique used in calculus to find the derivative of a function that is the ratio of two differentiable functions. It is particularly useful when you need to differentiate expressions of the form \( \frac{u}{v} \). In this exercise, we have a function \( y(t) = \frac{50}{1 + 6e^{-2t}} \). Here, the function can be seen as a fraction where the numerator \( u \) is a constant 50 and the denominator \( v \) is \( 1 + 6e^{-2t} \).

The quotient rule formula is:\[ \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]

• \( u \) here is a constant, so its derivative \( u' \) is 0.
• \( v = 1 + 6e^{-2t} \), so \( v' \) after differentiating is \(-12e^{-2t}\).

Substituting these into the quotient rule gives us the derivative \( \frac{dy}{dt} \) as \( \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \). This provides a crucial step for understanding how the slope of this curve changes.

Derivative Analysis

To fully understand the behavior of the function, we must analyze its derivative, which represents the slope of the curve. The derivative found using the quotient rule is \( \frac{600e^{-2t}}{(1 + 6e^{-2t})^2} \). This derivative shows us how the function \( y(t) = \frac{50}{1 + 6e^{-2t}} \) changes at any point \( t \).

Key Points to Note:

• The exponential term \( e^{-2t} \) means the derivative is always positive, indicating the function is increasing.
• To find where the curve steepens, it's vital to look at the behavior of this derivative.

By trying to find the maximum value of the derivative, we can identify the point at which \( y(t) \) climbs most steeply. This involves setting up and solving \( \frac{d^2y}{dt^2} = 0 \) to find potential points of maximum slope.
The analysis ensures we understand not just the change but the accelerate of that change, leading us to locate the curve’s steepest point accurately.

Curve Steepness

Curve steepness is determined where the derivative attains its maximum value. Simply put, it's where the curve climbs or descends the fastest. In scenarios like this one, finding the steepest point involves more than just calculating the first derivative.

Steps to Determine Steepness:

• Start by finding the first derivative, which we did using the quotient rule.
• Dive deeper by differentiating again: \( \frac{d^2y}{dt^2} \).
• Solve \( \frac{d^2y}{dt^2} = 0 \) to find critical points that might indicate local maximum or minimum slopes.
• Verify these points using the second derivative test to confirm they correspond to a maximum.

Once the critical point is pinpointed, evaluate it in the context of \( t \geq 0 \). Plugging the value of \( t \) back into the original function can give the \( y \) coordinate, confirming the steepest point's location on the curve. Understanding the steepness ensures clarity about how dynamically the curve behaves in its ascent or descent.