To solve this problem, we will apply related rates. Given that a right triangle has legs of 3 inches and 4 inches, and both sides are changing, we can use the Pythagorean theorem to relate the sides and their rates of change.

Step 1: Label the sides and variables

• Let $x$ be the length of the short leg (3 inches initially).
• Let $y$ be the length of the long leg (4 inches initially).
• Let $z$ be the length of the hypotenuse, which we need to find the rate of change for.

The Pythagorean theorem states: $x^2 + y^2 = z^2$ We are given:

• $\frac{dx}{dt} = 10$ inches per second (rate of change of the short leg).
• $\frac{dy}{dt} = 9$ inches per second (rate of change of the long leg).

We need to find $\frac{dz}{dt}$, the rate of change of the hypotenuse.

Step 2: Differentiate the Pythagorean theorem

Differentiating both sides of the equation $x^2 + y^2 = z^2$ with respect to time $t$ gives: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}$ We can cancel out the factor of 2 on all terms: $x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt}$

Step 3: Find the value of $z$

Using the Pythagorean theorem with the initial values of $x = 3$ and $y = 4$: $z^2 = 3^2 + 4^2 = 9 + 16 = 25$ $z = \sqrt{25} = 5$

Step 4: Substitute known values into the differentiated equation

Now, substitute the known values $x = 3$, $y = 4$, $z = 5$, $\frac{dx}{dt} = 10$, and $\frac{dy}{dt} = 9$ into the differentiated equation: $3(10) + 4(9) = 5 \frac{dz}{dt}$ $30 + 36 = 5 \frac{dz}{dt}$ $66 = 5 \frac{dz}{dt}$ $\frac{dz}{dt} = \frac{66}{5} = 13.2 \, \text{in/sec}$

Final Answer:

The rate of change of the hypotenuse is $13.2$ inches per second.

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Would you like more details on the steps or have any further questions? Here are some related questions:

1. How would the solution change if the rates of change of the legs were different?
2. Can this approach be applied to other geometric shapes with changing dimensions?
3. How do you differentiate implicitly when dealing with related rates in other contexts?
4. How does the rate of change of the area of a triangle relate to its side lengths changing?
5. What happens to the rate of change of the hypotenuse if the short leg's rate of change decreases?

Tip: When working with related rates, always start by writing down the equation that relates the quantities and then differentiate with respect to time.