Question

Once Kate's kite reaches a height of \(50 \mathrm{ft}\) (above her hands), it rises no higher but drifts due east in a wind blowing \(5 \mathrm{ft} / \mathrm{s} .\) How fast is the string running through Kate's hands at the moment that she has released \(120 \mathrm{ft}\) of string?

Short answer

Answer: The string is being released at a rate of 25/6 ft/s when there is 120 ft of string out.

Step-by-step solution

Define Variables and Write Down Given Information

We have the following information: - The height of the kite (b) is \(50 \mathrm{ft}\) - The kite drifts eastwards with a horizontal speed of \(5 \mathrm{ft} / \mathrm{s}\) - We want to find the speed that the string is being released when the length of the string (c) is \(120 \mathrm{ft}\)

Represent the Triangle Using the Pythagorean Theorem

To represent the relationship between the height, the eastward distance, and the length of the string, we can use the Pythagorean theorem: \(a^2 + b^2 = c^2\) Here, we know: - "a" is the distance drifted eastward - "b" is the height of the kite, which is \(50 \mathrm{ft}\) - "c" is the length of the string

Differentiate the Pythagorean Theorem with Respect to Time

Now, differentiate both sides of the equation with respect to time t: \(\frac{d}{dt}(a^2) + \frac{d}{dt}(b^2) = \frac{d}{dt}(c^2)\) Applying the chain rule, we obtain: \(2a\frac{da}{dt}+0 =2c\frac{dc}{dt}\) Since the height (b) is constant, its derivative is zero.

Use Given Values to Solve for dc/dt

We know the following values: - a = the horizontal distance drifted by the kite - \(b = 50 \mathrm{ft}\) - \(c = 120 \mathrm{ft}\) - \(\frac{da}{dt} = 5 \mathrm{ft} / \mathrm{s}\) (rate of eastward drift) Using the Pythagorean theorem, we can find the value of "a" when the string has a length of \(120 \mathrm{ft}\): \(a^2 + b^2 = c^2\) \(a^2 + (50^2) = (120^2)\) \(a^2 = (120^2) - (50^2)\) \(a = \sqrt{(120^2) - (50^2)} = 100 \mathrm{ft}\) Now let's substitute these values into the differentiated equation: \(2(100)\frac{da}{dt} =2(120)\frac{dc}{dt}\) \((200)(5)=2(120)\frac{dc}{dt}\)

Solve for the Rate of the String Being Released (dc/dt)

Finally, we need to find \(\frac{dc}{dt}\): \((200)(5)=2(120)\frac{dc}{dt}\) \(1000=240\frac{dc}{dt}\) \(\frac{dc}{dt} = \frac{1000}{240}= \frac{25}{6} \mathrm{ft}/\mathrm{s}\) So, the string is running through Kate's hands at a rate of \(\frac{25}{6} \mathrm{ft}/\mathrm{s}\) when she has released \(120 \mathrm{ft}\) of string.

Key concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental concept in geometry that relates the three sides of a right-angled triangle. It's given by the formula: \[ a^2 + b^2 = c^2 \]Where:

• \( a \) and \( b \) are the shorter sides, or "legs," of the triangle.
• \( c \) is the hypotenuse, the longest side opposite the right angle.

In the context of Kate's kite, the ground forms one leg \( a \) of the triangle, the constant height \( b = 50 \text{ ft} \) forms the other, and the string \( c \) acts as the hypotenuse. By applying the Pythagorean theorem, we determine the horizontal distance the kite has drifted when \( c = 120 \text{ ft} \). This theorem allows us to find unknown lengths, which are crucial for calculating rates of change, as shown later in other sections.

differentiation with respect to time

Differentiation with respect to time (denoted as \( \frac{d}{dt} \)) is crucial in related rates problems, where one or more quantities change over time. In our problem, we want to understand how the length of the string (a function of time) and other distances change with time. First, we apply differentiation to the equation from the Pythagorean theorem:\[ \frac{d}{dt}(a^2) + \frac{d}{dt}(b^2) = \frac{d}{dt}(c^2) \]This process turns static geometric relationships into dynamic equations that represent how these distances change over time, providing a pathway to solve for rates of change like the speed at which the string is released.

chain rule

The chain rule is an essential tool in calculus for differentiating expressions where one variable depends on another, which in turn depends on a third variable. In this problem, both \( a \) and \( c \) depend implicitly on time \( t \), so we use the chain rule during differentiation:

• For \( a^2 \), the derivative becomes \( 2a \cdot \frac{da}{dt} \)
• For \( c^2 \), it becomes \( 2c \cdot \frac{dc}{dt} \)

These results come from the general application of the chain rule: \( \frac{d}{dt} [u(t)^2] = 2u(t) \cdot \frac{du}{dt} \). By applying the chain rule, we're able to connect the changes in horizontal distance and the string length with the rate of eastward drift \( \frac{da}{dt} \) and the rate the string is released \( \frac{dc}{dt} \), respectively.

rate of change

The concept of rate of change is at the heart of related rates problems. It quantifies how fast one quantity changes in relation to another. In the kite problem, we are focused on determining how fast the string's length increases, denoted as \( \frac{dc}{dt} \). For each time increment, \( c \) increases due to the eastward wind, and \( da/dt = 5 \text{ ft/s} \) indicates the kite's horizontal drift rate. Using the differentiated equation and our solved values, we plug them in to find \( \frac{dc}{dt} \):\[ (200)(5) = 2(120) \cdot \frac{dc}{dt} \] These steps lead us to calculate the kite string running speed at \( \frac{25}{6}\text{ ft/s} \). This value tells us how quickly Kate's kite is letting out the string, pursuing a balanced understanding of real-time changes.