Algebra.Com's Answer #709095 by jim_thompson5910(35256)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
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document.write( "Part A)
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document.write( "Here's one way you can draw it out
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document.write( "![](\"https://i.imgur.com/QjYx8ZH.png\")
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document.write( "The points are defined as follows:
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document.write( "Point A = base of lamp
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document.write( "Point B = top of lamp
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document.write( "Point C = base of Bob's feet
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document.write( "Point D = top of Bob's head
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document.write( "Point E = edge of Bob's shadow (assuming the only light source is the lamp)\r
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document.write( "Based on those points and the given lengths, we know that
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document.write( "AB = 6 (lamp's height)
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document.write( "CD = 2 (Bob's height)
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document.write( "which are both in meters\r
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document.write( "We also have these unknowns
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document.write( "AC = x
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document.write( "CE = y
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document.write( "DE = z\r
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document.write( "Triangle BAE is similar to triangle DCE. So we can set up the proportion
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document.write( "AB/AE = CD/CE
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document.write( "AB/(AC+CE) = CD/CE
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document.write( "6/(x+y) = 2/y
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document.write( "6y = 2(x+y)
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document.write( "6y = 2x+2y
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document.write( "6y-2y = 2x+2y-2y
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document.write( "4y = 2x
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document.write( "4y/4 = 2x/4
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document.write( "y = x/2\r
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document.write( "So you have the correct answer for part A. Nice job.
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document.write( "Part B)\r
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document.write( "The first given bit of info states that \"Bob... walks away from a lamppost...at a constant rate of 1.5 meters per second\"\r
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document.write( "So this means that x is changing at a rate of 1.5 meters per second; therefore dx/dt = 1.5\r
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document.write( "dx/dt is the instantaneous speed at a given time t. In this case, it's always 1.5 as it is constant. The expression dx/dt is also positive because he is walking away from the lamp so x will be increasing as t increases.\r
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document.write( "The goal is to find the expression for dy/dt because we want to know how fast the shadow length is increasing at time t. The value y is the length of the shadow (see diagram in part A)\r
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document.write( "We'll use the result from part A to get us going. Simply differentiate both sides with respect to t\r
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document.write( "y = x/2
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document.write( "d/dt [ y ] = d/dt [ x/2 ]
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document.write( "dy/dt = (1/2) * d/dt [ x ]
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document.write( "dy/dt = (1/2) * (dx/dt)
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document.write( "dy/dt = (1/2) * (1.5)
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document.write( "dy/dt = 0.75\r
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document.write( "So the shadow is increasing at a constant rate of 0.75 meters per second
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document.write( "Part C)\r
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document.write( "Go back to the drawing in part A. Focus on triangle DCE. We have a right triangle with legs of 2 and y. The hypotenuse is z.\r
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document.write( "Using the Pythagorean Theorem, we can say,
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document.write( "a^2 + b^2 = c^2
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document.write( "2^2 + y^2 = z^2
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document.write( "4 + y^2 = z^2\r
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document.write( "Note: we can solve for z to get 
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document.write( "However, it is easier to stick with 4 + y^2 = z^2 when it comes to implicit differentiation. \r
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document.write( "Part D)\r
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document.write( "If Bob is 7 meters from the lamp, then x = 7. So y is
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document.write( "y = x/2
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document.write( "y = 7/2
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document.write( "y = 3.5\r
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document.write( "Then we can figure out z based on y = 3.5
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document.write( "4 + y^2 = z^2
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document.write( "4 + (3.5)^2 = z^2
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document.write( "16.25 = z^2
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document.write( "z^2 = 16.25
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document.write( "sqrt(z^2) = sqrt(16.25)
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document.write( "z = 4.03112887 (this is approximate)\r
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document.write( "We'll use the result from part C. Now implicitly differentiate both sides with respect to t
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document.write( "4 + y^2 = z^2
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document.write( "d/dt[4 + y^2] = d/dt[z^2]
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document.write( "d/dt[4] + d/dt[y^2] = d/dt[z^2]
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document.write( "0 + 2y*d/dt[y] = 2z*d/dt[z]
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document.write( "0 + 2y*dy/dt = 2z*dz/dt
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document.write( "2y*dy/dt = 2z*dz/dt\r
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document.write( "Plug in the values found earlier; isolate dz/dt
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document.write( "2y*dy/dt = 2z*dz/dt
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document.write( "2*(3.5)*(0.75) = 2*4.03112887*dz/dt
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document.write( "5.25 = 8.06225774*dz/dt
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document.write( "dz/dt = 5.25/8.06225774
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document.write( "dz/dt = 0.65118236 (this is approximate)\r
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document.write( "So the distance from Bob's head to the tip of the shadow is increasing at roughly 0.65118236 meters per second at the exact moment when he is 7 meters from the lamp. \r
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document.write( "Unlike dx/dt = 1.5 and dy/dt = 0.75, the value of dz/dt is not constant. It depends on y (which is ultimately dependent on x).\r
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