Question

A ladder $10$ m long against a vertical wall with the lower and on the horizontal ground.

The lower end of the ladder is pulled along the ground away from the wall at the rate of $3$cm/sec.

The height of the upper end while it is descending at the rate of $4$ cm/sec is

• A. $4\sqrt{3}$ m
• B. $5\sqrt{3}$ m
• C. $5\sqrt{2}$ m
• D. $8$ m
• E. $6$ m

Answer 1

The correct option is E

$6$ m

The explanation for the correct option:

Step 1: Calculate the third side

Since the given triangle is a right-angle triangle

Using the Pythagorean theorem,

$$\begin{gathered} x^2+y^2={10}^2 \\ {x}^2+y^2=100 \end{gathered}$$

Now,

$$\begin{gathered} \frac{dx}{dt}=3cm/sec \\ \frac{dy}{dt}=4cm/sec \end{gathered}$$

Step 2: Finding the values of x and y

$\begin{array}{rcl}2x\left(\frac{dx}{dt}\right)+2y\left(\frac{dy}{dt}\right)& =& 0\\ x\left(\frac{dx}{dt}\right)+y\left(\frac{dy}{dt}\right)& =& 0\\ 3x–4y& =& 0\\ 4y& =& 3x\\ x& =& \frac{4y}{3}\end{array}$

Step 3. Put the value of $x$ in $x^2+y^2=100$

$\begin{array}{rcl}{x}^2+y^2& =& 100\\ {\left(\frac{4y}{3}\right)}^2+y^2& =& 100\\ 16y^2+9y^2& =& 900\\ 25y^2& =& 900\\ y^2& =& 36\\ y& =& 6\end{array}$

Hence, the height of the upper end is 6 m