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Solve the problems in related rates.A light in a garage is $9.50 \mathrm{ft}$ above the floor and $12.0 \mathrm{ft}$ behind the door. If the garage door descends vertically at $1.50 \mathrm{ft} / \mathrm{s}$, how fast is the door's shadow moving toward the garage when the door is 2.00 ft above the floor?
Calculus 1 / AB
Chapter 24
Applications of the Derivative
Section 4
Related Rates
Derivatives
Missouri State University
University of Michigan - Ann Arbor
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okay, We know that we're looking for the height of the shadow on the wall. H is a function of the distance. So by similarity of triangles, actually, they suggested sketching in the problem. So perhaps we should do that. So if we've got a stick figure person over here, that's five feet. We know the distance from here to here is 10 feet and the total distance from the end is 40 feet. And then obviously you've got the wall. Okay, So by the similarities of triangles, this is from geometry. We've got h over 40 is five over 40 minus x rate 40 minus x. So this gives us a TSH equals 200 over 14 minus x. If we take the derivative of this, we end up with D H. Over DT is 200 over 40 minutes. X squared times d axe over d t. Giving us de age over. DT is night of 4/9 because we know the person's walking at a rate of two feet per second towards the wall and we know the distance X is decreasing, has a rate of two feet per second. That's negative to write because of its decreasing its negative so negative for ninth industrial form is negative 0.44 feet per second
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