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Anna35 [415]
3 years ago
14

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
Mathematics
1 answer:
Alex777 [14]3 years ago
5 0

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:  

H^{2} = B^{2} + L^{2}

L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}

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

A = \frac{BL}{2}  

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

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

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

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

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

\frac{dA}{dt} = 15.75 ft^{2}/s  

     

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

I hope it helps you!                                    

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General Idea:

(i) Assign variable for the unknown that we need to find

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(iii) Write the mathematical equation representing the description given.

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Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

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<u>Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."</u>

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<u>Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."</u>

Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\

Conclusion:

The systems of equations that can be used to determine the length and width of the pen and the area of the doghouse is given in Option B.

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