35. The diagonal \( z \) of the rectangle at the right is increasing at the rate of 2 cm/sec and

\[ \frac{dy}{dt} = 3 \frac{dx}{dt} \]

At what rate is the length \( x \) increasing when \( x = 3 \) cm and \( y = 4 \) cm?

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Answer is: option3

\( \frac{2}{3} \) cm/sec

Solution:

We are given a right triangle where the diagonal \( z \) is increasing at a rate of:

\[ \frac{dz}{dt} = 2 \text{ cm/sec}. \]

We also have the relationship:

\[ \frac{dy}{dt} = 3 \frac{dx}{dt} \]

We need to determine \( \frac{dx}{dt} \) when \( x = 3 \) cm and \( y = 4 \) cm.

Step 1: Pythagorean Theorem

The diagonal \( z \) of the rectangle follows:

\[ z^2 = x^2 + y^2 \]

Differentiating both sides with respect to \( t \):

\[ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]

Dividing by 2:

\[ z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \]

Step 2: Substituting Given Values

At \( x = 3 \) cm and \( y = 4 \) cm, we use the Pythagorean theorem:

\[ z = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

We substitute \( z = 5 \), \( \frac{dz}{dt} = 2 \), \( x = 3 \), and \( y = 4 \):

\[ 5(2) = 3 \frac{dx}{dt} + 4 \left( 3 \frac{dx}{dt} \right) \]

Since we know \( \frac{dy}{dt} = 3 \frac{dx}{dt} \), substitute this:

\[ 10 = 3 \frac{dx}{dt} + 12 \frac{dx}{dt} \] \[ 10 = 15 \frac{dx}{dt} \] \[ \frac{dx}{dt} = \frac{10}{15} = \frac{2}{3} \text{ cm/sec} \]

Final Answer: \( \frac{2}{3} \) cm/sec.

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