As Collin walks away from a 264 cm lamppost, the tip of his shadow moves twice as fast as he does? What is Collin's height?
How should I do this?
The moment that a rectangle is 5 ft long and 4 ft wide, its length is increasing at 1 ft/min and its width is decreasing at 0.5 ft/min . How is the area changing at that moment? A= l*w , Da/dt = w*dw/dt +l*dl/dt =4*1(-0.5 ft/min)+ 5(1 ft/min) = 3 ft/min Da/dt = 3 ft/min is this correct?
Let
h = Colin's height
x = Colin's distance from post
s = distance of tip of shadow from post
(s-x)/h = s/264
264s -264x = hs
(264-h)s = 264x
s = 264/(264-h) x
ds/dt = 264/(264-h) dx/dt
since ds/dt = 2 dx/rt,
2 = 264/(264-h)
h = 132
To solve this problem, you need to set up a proportion based on the given information.
Let's say Collin's height is "h" cm.
According to the problem, as Collin walks away from the lamppost, the tip of his shadow moves twice as fast as he does. This means that for every one unit Collin walks, his shadow moves two units.
We can set up the following proportion:
h/264 = (2h)/(2*264)
Simplifying the proportion:
h/264 = h/264
This means that Collin's height is equal to the distance between him and the lamppost.
So, Collin's height is 264 cm.
To solve the first question, we can set up a proportion based on the relationship between Collin's height, the length of the lamppost, and the ratio of the movement of Collin's shadow to Collin's own movement.
Let's assume Collin's height is represented by 'h'. According to the given information, as Collin walks away from the lamppost, the tip of his shadow moves twice as fast as he does. This means that the rate of change of the shadow's length is twice the rate of change of Collin's height.
We can set up the following proportion:
(h / (264 - h)) = 2
By cross-multiplying and simplifying, we get:
2h = 264 - h
Solving for 'h', we find:
3h = 264
h = 88 cm
Therefore, Collin's height is 88 cm.
For the second question, it seems that you have correctly applied the formula for finding the rate of change of the area of a rectangle. The formula you used is:
Da/dt = w * dw/dt + l * dl/dt
Where 'Da/dt' represents the rate of change of the area, 'w' is the width, 'dw/dt' is the rate of change of the width, 'l' is the length, and 'dl/dt' is the rate of change of the length.
Substituting the given values:
Da/dt = 4 * 1 (-0.5 ft/min) + 5 * (1 ft/min)
Da/dt = -2 ft²/min + 5 ft²/min
Da/dt = 3 ft²/min
Therefore, you are correct. The rate of change of the area at that moment is 3 ft²/min.