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Applications of differentiation?
A 2-foot tall dog is walking away from a streetlight which is on a 10-foot pole. At a certain moment, the tip of the dog's shadow is moving away from the streetlight at 5 feet per second. How fast is the dog walking at that moment?
Explain, please!
1 Answer
- Anonymous7 years ago
Let the distance of the streetlight from the dog = x
Let the length of the shadow from the dog = y.
The speed of the dog is dx / dt
The speed of the shadow is d(x + y) / dt
Now, using similar triangles: (Draw the triangle, you should be able to see this)
(x + y) / 10 = y / 2
2x + 2y = 10y
y = ¼x
Remember the speed of the dog's shadow:
d(x + y) / dt = 5
d(x + ¼x) / dt = 5
d[(5/4)x] / dt = 5 --- {By constant multiple rule, 5/4 comes out front)
(5/4)dx/dt = 5
dx/dt = 4
Remember that dx/dt represents the speed of the dog.