Problem 29 - Calculus Single Variable

Question

A container is pulled up a \(10 \mathrm{ft}\) long ramp from the ground to a loading dock \(4 \mathrm{ft}\) above the ground. At the moment it is halfway along the ramp, it is moving at the rate of \(1.2 \mathrm{ft} / \mathrm{s}\). How fast is it rising at that instant?

Step-by-Step Solution

Answer

The container is rising at approximately 0.69 ft/s vertically when halfway up the ramp.

Step 1: Understand the Problem

We need to find out how fast the container is rising vertically when it is halfway along a 10-foot ramp. The rate at which it moves along this ramp at that moment is 1.2 feet per second.

Step 2: Sketch and Define Variables

Sketch a right triangle where the ramp is the hypotenuse. Let the horizontal distance along the ground be \(x\) and the vertical height be \(h\). At halfway, the distance along the ramp is \(s = 5 \) feet. The proportion gives \(x = \frac{5 \times 8}{10} = 4 \) feet.

Step 3: Use Pythagorean Theorem

Use the Pythagorean theorem for a right triangle: \[ x^2 + h^2 = 10^2 \]Plug in \( x = 5 \):\[ 5^2 + h^2 = 10^2 \]\[ 25 + h^2 = 100 \]\[ h^2 = 75 \]Solve for \(h\) when \(s = 5\), we need \(h\) just for rates; it was asked for midpoint initial horizontal position.

Step 4: Differentiate to Find Rates

Differentiate both sides with respect to time \(t\) to find \(\frac{dh}{dt}\):\[ 2x \frac{dx}{dt} + 2h \frac{dh}{dt} = 0 \]\[ x \frac{dx}{dt} + h \frac{dh}{dt} = 0 \]

Step 5: Solve for Vertical Rate of Change

Using the derivative equation:\[ h \frac{dh}{dt} = - x \frac{dx}{dt} \]Rearrange to find \( \frac{dh}{dt} \):\[ \frac{dh}{dt} = - \frac{x}{h} \times \frac{dx}{dt} \]Substitute \(\frac{dx}{dt} = 1.2\), and calculate \(h = \sqrt{75} \approx 8.66\):\[ \frac{dh}{dt} = - \frac{5}{8.66} \times 1.2 \approx - 0.69 \, \mathrm{ft/s} \]

Step 6: Interpret the Result

The negative sign indicates the vertical height is not actually decreasing but your error in direction is corrected, the faster vertical movement contributes positively.

Key Concepts

Pythagorean theorem

The Pythagorean Theorem is a key concept that helps in solving problems involving right triangles. It states that for any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In mathematical terms, this is expressed as:

• \( c^2 = a^2 + b^2 \)

In the exercise, the ramp forms the hypotenuse of a right triangle with the horizontal and vertical sides. When the container is halfway up the ramp, we use the theorem to find the vertical rise. This gives:

• \( 10^2 = x^2 + h^2 \)

Plugging in our information at the midpoint, we solve for the remaining distance, which originally helps set up our equation for finding rates of change later.

differentiation

Differentiation is a fundamental calculus method used to determine the rate at which a quantity changes. It's especially useful in related rates problems, where two or more quantities are related and change over time. In the given exercise, we're looking for the rate of rise (vertical rate) of the container.
To find how fast the height changes, we differentiate the Pythagorean equation, now with respect to time:

• \( 2x \frac{dx}{dt} + 2h \frac{dh}{dt} = 0 \)

Here, \( \frac{dx}{dt} \) is the rate of change of the horizontal distance. By differentiating, we relate this rate to \( \frac{dh}{dt} \), the vertical rise rate we're interested in.
This method integrates understanding of the relationships within the triangle to find the desired rate, considering how one dimension's change impacts another.

vertical rise rate

The vertical rise rate tells us how quickly the container is ascending vertically as it moves along the ramp. After determining the necessary derivative relationships in the differentiation step, we can solve for this specific rate.
From our differentiated equation from before:

• \( \frac{dh}{dt} = -\frac{x}{h} \times \frac{dx}{dt} \)

We substitute our known values into the equation: with \( x = 5 \), \( h = \sqrt{75} \approx 8.66 \), and \( \frac{dx}{dt} = 1.2 \) feet per second. These substitutions allow us to calculate \( \frac{dh}{dt} \):

• \( \frac{dh}{dt} = - \frac{5}{8.66} \times 1.2 \approx - 0.69 \text{ ft/s} \)

The negative sign in this solution indicates direction. Since we're finding how fast the container rises, the actual vertical velocity is positive—reflecting an upward movement.