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elopezhilario6339
07.10.2020 •
Mathematics
A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet of the ground (see figure).
(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?
Thank you in advance!
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Ответ:
a) ds/dt = 25/3 ft/s
b) 10/3 ft/s
Step-by-step explanation:
Using the triangles and the ratios given
15 ft s
=
6 ft s-x
where s is the length from the lamppost to the shadow and x is the length from the lamppost to the man
Using cross products
15( s-x) = 6s
15s -15x = 6s
15s-6s = 15x
9s = 15x
s = 15/9 x
s = 5/3 x
Taking the derivative of each side with respect to time
ds/dt = 5/3 dx/dt
We know dx/dt = 5 ft/s
ds /dt = 5/3 * 5 = 25/3 ft/s
a) ds/dt = 25/3 ft/s
The length of the shadow is ( s-x)
d/dt ( s-x)
ds/dt - dx/dt
25/3 - 5
25/3 - 15/3
10/3 ft/s
b) 10/3 ft/s
Ответ:
$138.24
Step-by-step explanation: