A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 9 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 6 feet from the wall

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s        

First, we need to find the other side of the triangle:  

[tex]H^{2} = B^{2} + L^{2}[/tex]

[tex] L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}} [/tex]

Now, the area (A) of the triangle is:            

[tex] A = \frac{BL}{2} [/tex]  

Hence, the rate of change of the area is given by:

[tex] \frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}] [/tex]      

[tex] \frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}] [/tex]        

[tex]\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}][/tex]  

[tex]\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9][/tex]      

[tex]\frac{dA}{dt} = 15.75 ft^{2}/s[/tex]  

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!                                    

The rate of change (ROC) of the area with respect to (w.r.t.) time can be

found from the ROC of the area w.r.t. x and the ROC of x w.r.t. time.

Reasons:

The length pf the ladder = 10 feet

Rate at which the ladder is pulled from the wall, [tex]\displaystyle \frac{dx}{dt}[/tex] = 9 feet per second

Required:

The rate at which the area of the triangle formed by the ladder, the wall

and the ground, is changing at the instant the ladder is 6 feet from the wall.

Solution:

The area the triangle, A = 0.5·x·y

Where;

x = The distance of the ladder from the wall

y = The height of the ladder on the wall

By Pythagoras's theorem, we have;

10² = x² + y²

Which gives;

y = √(10² - x²)

Therefore;

The area the triangle, A = 0.5 × x × √(10² - x²)

By chain rule, we have;

[tex]\displaystyle \frac{dA}{dt} = \mathbf{\frac{dA}{dx} \times \frac{dx}{dt}}[/tex]

[tex]\displaystyle \frac{dA}{dx} = \frac{d\left(0.5 \cdot x \cdot \sqrt{10^2 - x^2} }{dx} = \mathbf{\frac{\left(x^2 - 50\right) \cdot \sqrt{100-x^2} }{x^2-100}}[/tex]

[tex]\displaystyle \frac{dA}{dx} = \frac{\left(x^2 - 50\right) \cdot \sqrt{100-x^2} }{x^2-100}[/tex]

Therefore;

[tex]\displaystyle \frac{dA}{dt} = \mathbf{\frac{\left(x^2 - 50\right) \cdot \sqrt{100-x^2} }{x^2-100} \times 9}[/tex]

When the ladder is 6 feet from the wall, we have;

x = 6

[tex]\displaystyle \frac{dA}{dt} = \frac{\left(6^2 - 50\right) \cdot \sqrt{100-6^2} }{6^2-100} \times 9 = \mathbf{15.75}[/tex]

At the time the ladder is 6 feet from the wall, the area is increasing at 15.75 ft.²/sec.

Learn more about chain rule of differentiation here:

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