Problem 32 - University Calculus: Early Transcendentals

Question

vent in the crater's floor, which at one point shot lava \(1900 \mathrm{ft}\) straight into the air (a Hawaiian record). What was the lava's exit velocity in feet per second? In miles per hour? (Hint: If \(v_{0}\) is the exit velocity of a particle of lava, its height \(t\) sec later will be \(s=v_{0} t-16 t^{2} \mathrm{ft} .\) Begin by finding the time at which \(d s / d t=0\) Neglect air resistance.) Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single

Step-by-Step Solution

Answer

The exit velocity is 348.3 ft/s or approximately 237.4 mph.

Step 1: Understand the Problem

We are given an equation for the height of lava based on time and exit velocity: \( s = v_0 t - 16t^2 \). We need to find the initial velocity \( v_0 \) when the lava reaches its maximum height, which is 1900 feet. At maximum height, the derivative \( \frac{ds}{dt} = 0 \).

Step 2: Differentiate the Height Equation

Differentiate the equation \( s = v_0 t - 16t^2 \) with respect to \( t \) to find \( \frac{ds}{dt} \). This gives \( \frac{ds}{dt} = v_0 - 32t \).

Step 3: Find the Time at Maximum Height

Set the derivative equal to zero to find the time when the lava is at its maximum height: \( v_0 - 32t = 0 \). This simplifies to \( t = \frac{v_0}{32} \).

Step 4: Substitute Time into the Height Equation

Substitute \( t = \frac{v_0}{32} \) into the original height equation: \( 1900 = v_0 \cdot \frac{v_0}{32} - 16 \cdot \left(\frac{v_0}{32}\right)^2 \).

Step 5: Solve for Exit Velocity

Simplify and solve for \( v_0 \): \[ 1900 = \frac{v_0^2}{32} - \frac{v_0^2}{64} = \frac{v_0^2}{64} \Rightarrow v_0^2 = 1900 \times 64 = 121600 \Rightarrow v_0 = 348.3 \text{ ft/s} \].

Step 6: Convert Feet per Second to Miles per Hour

Convert the velocity from feet per second to miles per hour: \( 348.3 \text{ ft/s} \times \frac{1 \text{ mile}}{5280 \text{ ft}} \times 3600 \text{ s/hr} \approx 237.4 \text{ mph} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, mainly used to find the rate at which a quantity changes. In this exercise, we apply differentiation to find the velocity of the lava as it reaches its maximum height.

To start, we have the equation for the lava height: \( s = v_0 t - 16t^2 \). The term \( v_0 t \) represents the lava's initial speed as a factor of time, and \( -16t^2 \) is a constant that accounts for gravitational acceleration downwards.

• Differentiate the height equation \( s \) with respect to time \( t \) to get the rate of change of height, \( \frac{ds}{dt} \).
• Using basic differentiation rules, we find \( \frac{ds}{dt} = v_0 - 32t \).
• This equation represents how the height of the lava changes over time and helps us understand when the lava stops rising (i.e., when\( \frac{ds}{dt} = 0 \)).

Maximum Height

The maximum height of the lava is a key point where its velocity becomes zero momentarily as it changes direction. This is critical because at maximum height, all kinetic energy is converted into potential energy.

The process to find the maximum height involves setting the derivative \( \frac{ds}{dt} = 0 \). This tells us when the lava reaches the peak of its trajectory.

• Start by setting \( v_0 - 32t = 0 \) which yields the time \( t = \frac{v_0}{32} \).
• This time \( t \) is when the lava reaches its highest point above the ground.
• By substituting this \( t \) back into the original height equation \( s \), we can understand how high the lava actually goes, which has been given as 1900 ft.

Velocity Calculation

Velocity calculation is about determining the speed of the lava at its initial exit. This is important to understand the energy and power of volcanic eruptions.

Once we've set up the equation for when the lava reaches its maximum height, we solve for \( v_0 \) using the height equation.

• Initially, we found that \( t = \frac{v_0}{32} \).
• Substitute this time back into the height equation: \( 1900 = v_0 \cdot \frac{v_0}{32} - 16 \cdot \left(\frac{v_0}{32}\right)^2 \).
• Simplify to solve for \( v_0^2 \), leading to \( v_0 = 348.3 \text{ ft/s} \).

This velocity precisely estimates the initial speed at which the lava was expelled from the vent.

Unit Conversion

Unit conversion allows us to express the same physical quantity in different units, making it easier to understand and compare. Here, the initial velocity was given in feet per second, but is often more comprehensible in miles per hour.

To convert from feet per second (ft/s) to miles per hour (mph), we perform a few basic steps:

• There are 5280 feet in a mile and 3600 seconds in an hour. Use these conversions to switch units.
• Calculate \( 348.3 \text{ ft/s} \times \frac{1 \text{ mile}}{5280 \text{ ft}} \times 3600 \text{ s/hr} \approx 237.4 \text{ mph} \).
• This conversion shows that the lava's exit velocity is around 237.4 mph, illustrating the immense force involved in volcanic eruptions.

Thus, converting units ensures that we can communicate our findings effectively in different contexts.