Respuesta :
As the ladder is pulled away from the wall, the area and the height with the
wall are decreasing while the angle formed with the wall increases.
The correct response are;
- (a) The velocity of the top of the ladder = 1.5 m/s downwards
- (b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec
- (c) The rate at which the angle formed with the wall is changing is approximately 0.286 rad/sec.
Reasons:
The given parameter are;
Length of the ladder, l = 25 feet
Rate at which the base of the ladder is pulled, [tex]\displaystyle \frac{dx}{dt}[/tex] = 2 feet per second
(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;
[tex]\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}[/tex]
25² = x² + y²
y = √(25² - x²)
[tex]\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}[/tex]
Which gives;
[tex]\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2[/tex]
[tex]\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}[/tex]
When x = 15, we get;
[tex]\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}[/tex]
The velocity of the top of the ladder = 1.5 m/s downwards
When x = 20, we get;
[tex]\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6[/tex]
The velocity of the top of the ladder = [tex]\underline{-2.\overline{6} \ m/s \ downwards}[/tex]
When x = 24, we get;
[tex]\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86[/tex]
The velocity of the top of the ladder ≈ -6.86 m/s downwards
(b) [tex]\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}[/tex]
Therefore;
[tex]\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}[/tex]
[tex]\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}[/tex]
[tex]\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}[/tex]
Therefore;
[tex]\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2[/tex]
When the ladder is 24 feet from the wall, we have;
x = 24
[tex]\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}[/tex]
The rate the area formed by the ladder is changing, [tex]\displaystyle \frac{dA}{dt}[/tex] ≈ -75.29 ft.²/sec
(c) From trigonometric ratios, we have;
[tex]\displaystyle sin(\theta) = \frac{x}{25}[/tex]
[tex]\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}[/tex]
[tex]\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}[/tex]
[tex]\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}[/tex]
Which gives;
[tex]\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}[/tex]
When x = 24 feet, we have;
[tex]\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}[/tex]
Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is [tex]\displaystyle \frac{d \theta}{dt}[/tex] ≈ 0.286 rad/sec
Learn more about the chain rule of differentiation here:
https://brainly.com/question/20433457
