Anna is 6 ft. tall. She is walking away from a street light that is 24 ft tall at a rate of 4 ft/sec. How fast is the length of her shadow changing?
Using similar triangles
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To find the rate at which the length of Anna's shadow is changing, you can use related rates and similar triangles.
Let (x) be the distance between Anna and the base of her shadow, and let (y) be the length of her shadow.
Using similar triangles, we have: [\frac{x}{24} = \frac{x+y}{6}]
Differentiating both sides with respect to time, we get: [\frac{d}{dt}\left(\frac{x}{24}\right) = \frac{d}{dt}\left(\frac{x+y}{6}\right)]
Given that ( \frac{dx}{dt} = -4 ) ft/sec (because Anna is moving away from the light), we need to find ( \frac{dy}{dt} ).
Differentiating both sides with respect to ( t ), we get: [\frac{1}{24}\frac{dx}{dt} = \frac{1}{6}\left(\frac{dx}{dt} + \frac{dy}{dt}\right)]
Substitute ( \frac{dx}{dt} = -4 ) into the equation and solve for ( \frac{dy}{dt} ):
[\frac{1}{24}(-4) = \frac{1}{6}(-4 + \frac{dy}{dt})]
[-\frac{1}{6} = -\frac{4}{6} + \frac{dy}{6}]
[\frac{dy}{dt} = -\frac{1}{6}]
So, the length of Anna's shadow is changing at a rate of ( -\frac{1}{6} ) ft/sec.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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