AP Calculus AB / Contextual Applications of Differentiation / Question No. 35

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

\frac{d y}{d t} = 3 \frac{d x}{d t}

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

1 cm/sec

\frac{3}{4}

cm/sec

\frac{2}{3}

cm/sec

\frac{1}{3}

cm/sec

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{d z}{d t} = 2 cm/sec .

We also have the relationship:

\frac{d y}{d t} = 3 \frac{d x}{d t}

We need to determine $\frac{d x}{d t}$ 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 $t$ :

2 z \frac{d z}{d t} = 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t}

Dividing by 2:

z \frac{d z}{d t} = x \frac{d x}{d t} + y \frac{d y}{d t}

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 $z = 5$ , $\frac{d z}{d t} = 2$ , $x = 3$ , and $y = 4$ :

5 (2) = 3 \frac{d x}{d t} + 4 (3 \frac{d x}{d t})

Since we know $\frac{d y}{d t} = 3 \frac{d x}{d t}$ , substitute this:

10 = 3 \frac{d x}{d t} + 12 \frac{d x}{d t}

10 = 15 \frac{d x}{d t}

\frac{d x}{d t} = \frac{10}{15} = \frac{2}{3} cm/sec

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

Previous Next