A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. (a) When he is 10 feet from the base of the light, at what rate is the tip of his shadow moving? (b) When he is 10 feet from the base of the light, at what rate is the length of his shadow changing?

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pseudoenergy34 · Accepted Answer

Look at the ilustration in the book.  The lamppost is 15 ft. tall, and the man is 6 ft. tall. We use a proportion to find the shadow. s is the length of the base of the lamppost to the shadow, x is the length of the base of the lamppost to the man, so the length of the shadow is s - x. The lamppost is 15 ft. tall, and the man is 6 ft. tall. We use a proportion to find the shadow. s is the length of the base of the lamppost to the shadow, x is the length of the base of the lamppost to the man, so the length of the shadow is s - x.  156=ss−x  Cross multiply and distribute  15s−15x=6s  Isolate the variable s, and simplify the fraction.  s=53x  Take the derivative according to time.  dsdt=5dx3dt  Given that dxdt is 5 ft/sec, multply. This answer for a.  dsdt=53×5=253 ft/sec  dsdt we found in part a, dxdt is given in the problem.  dsdt=253,dxdt=5  Subtract. This is the answer to part b.  253−5=103 ft/sec

star233 · Accepted Answer

To solve the given problem, we can use similar triangles and apply the chain rule from calculus. Let&#039;s proceed with the solution:(a) Let&#039;s assume that the length of the man&#039;s shadow is represented by s and the distance between the man and the light is represented by x. We need to find the rate at which the tip of his shadow is moving, which is the derivative dsdt.By similar triangles, we can set up the following proportion:{length&amp;nbsp;of&amp;nbsp;shadow}{height&amp;nbsp;of&amp;nbsp;man}={distance&amp;nbsp;from&amp;nbsp;light+{length&amp;nbsp;of&amp;nbsp;shadow}}{height&amp;nbsp;of&amp;nbsp;light}Using the given values, we have:s6=x+s15Now, let&#039;s solve this equation for s:15s=6(x+s)15s=6x+6s9s=6xs=2x3To find dsdt, we can take the derivative of s with respect to time t. Since x is changing with time, we need to use the chain rule:dsdt=dsdx·dxdtTo find dsdx, we differentiate the equation s=2x3 with respect to x:dsdx=23Given that the man is moving away from the light at a rate of 5 feet per second (dxdt=5), we can substitute the values into the equation:dsdt=23·5dsdt=103Therefore, when the man is 10 feet from the base of the light, the tip of his shadow is moving at a rate of 103 feet per second.(b) To find the rate at which the length of his shadow is changing, we need to find dℓdt, where ℓ represents the length of the shadow.Using the Pythagorean theorem, we can write:ℓ2=s2+x2Differentiating both sides with respect to time t, we have:2ℓ·dℓdt=2s·dsdt+2x·dxdtSubstituting the known values, we get:2ℓ·dℓdt=2s·dsdt+2x·dxdt2ℓ·dℓdt=2(2x3)·103+2x·52ℓ·dℓdt=40x9+10x2ℓ·dℓdt=40x9+10x2ℓ·dℓdt=40x+90x92ℓ·dℓdt=130x9Simplifying the equation, we have:ℓ·dℓdt=130x18Now, we know that when the man is 10 feet from the base of the light, the length of his shadow is given by ℓ=s2+x2. Substituting the values, we get:ℓ=(2x3)2+x2ℓ=4x29+x2ℓ=13x29ℓ=13x3Now, let&#039;s substitute this expression for ℓ into the equation we obtained earlier:(13x3)·dℓdt=130x18Simplifying further, we have:13x·dℓdt=130x18Dividing both sides by x and simplifying:13·dℓdt=13018Finally, solving for dℓdt, we find:dℓdt=1301813Therefore, when the man is 10 feet from the base of the light, the length of his shadow is changing at a rate of 1301813 feet per second.

alenahelenash · Accepted Answer

Let&#039;s denote the height of the man as h, the distance from the man to the base of the light as x, the length of the shadow as s, and the rate at which the man is walking as dxdt.(a) We want to find the rate at which the tip of the shadow is moving, which is dsdt.From the similar triangles formed by the man, his shadow, and the light, we have the following relationship:hs=h+15xDifferentiating both sides of the equation with respect to time t, we get:dhdt·1s−hs2·dsdt=dhdt·1x−h+15x2·dxdtWe are given that dhdt=0 (since the man&#039;s height doesn&#039;t change), dxdt=5 feet per second, h=6 feet, and x=10 feet.Plugging in these values, we can solve for dsdt:0·1s−6s2·dsdt=0·110−6+15102·5Simplifying the equation:−6s2·dsdt=−2120Now, solving for dsdt:dsdt=2120·s26Plugging in s=10 (since the length of the shadow is 10 feet when the man is 10 feet away from the base of the light), we can find the rate at which the tip of his shadow is moving:dsdt=2120·1026Therefore, the tip of his shadow is moving at a rate of 352 feet per second.(b) We want to find the rate at which the length of his shadow is changing, which is dsdt.From the similar triangles formed by the man, his shadow, and the light, we have the following relationship:h+15s=hxDifferentiating both sides of the equation with respect to time t, we get:dhdt·1s−h+15s2·dsdt=dhdt·1x−hx2·dxdtPlugging in the known values, we get:dsdt=hx·dhdt−h+15s·dsdtSubstituting the given values dhdt=0, dxdt=5 ft/s, h=6 ft, x=10 ft, we can solve for dsdt:0·1s−6+15s2·dsdt=0·110−610·5Simplifying the equation:−21s2·dsdt=−3Solving for dsdt:dsdt=37·s21Plugging in s=10, we can find the rate at which the length of his shadow is changing:dsdt=37·102Therefore, the length of his shadow is changing at a rate of 3007 ft/s.

xleb123 · Accepted Answer

Result:(a)13(b)415Solution:(a) Let x be the distance from the man to the base of the light, and let h be the length of the man&#039;s shadow. The height of the man&#039;s shadow is 15 feet, and the height of the man is 6 feet.Using similar triangles, we have xh=615.Differentiating both sides with respect to time t, we get dxdt·1h=615·dhdt.Since the man is walking away from the light source, dxdt=5 feet per second.We are interested in finding dhdt when x=10 feet.Substituting the given values, we have 5·1h=615·dhdt.Simplifying, we get dhdt=5h·615.At x=10 feet, we can find h using the proportion: 10h=615.Solving for h, we have h=756 feet.Substituting h into the expression for dhdt, we get dhdt=5756·615.Simplifying further, we find dhdt=13 feet per second.Therefore, the tip of the man&#039;s shadow is moving at a rate of 13 feet per second.(b) To find the rate at which the length of the man&#039;s shadow is changing, we differentiate the equation xh=615 with respect to time t.ddt(xh)=ddt(615).Using the quotient rule, we have 1h·dxdt−xh2·dhdt=0.Since dxdt=5 feet per second, we can substitute the known values and solve for dhdt.1756·5−10(756)2·dhdt=0.Simplifying, we find dhdt=5756·6(756)2.Further simplification yields dhdt=415 feet per second.Therefore, the length of the man&#039;s shadow is changing at a rate of 415 feet per second when he is 10 feet from the base of the light.

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is

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