A woman is sitting on top of a vertical cliff 200 feet
above a sea. She watches as a boat sails away at a rate
of 32 feet per second. When the boat is about 100 feet
from the base of the cliff, determine the rate at which
the angle of depression of the boat from the woman’s
viewpoint changes.
above a sea. She watches as a boat sails away at a rate
of 32 feet per second. When the boat is about 100 feet
from the base of the cliff, determine the rate at which
the angle of depression of the boat from the woman’s
viewpoint changes.
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For this question a diagram is necessary to visualize:
A______________________
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B
Here A is the position of the woman. AB is the cliff of height 200 ft. AB=200. C is the location of the boat at any time. BC=x increases as the boat moves away from the cliff at 32 ft/s. The angle of depression 'z' is the angle made by AC with the horizontal at A.
By alternate angles theorem angle ACB=z
tan z=AB/BC=200/x
differentiating both sides partially,
sec^2(z)*dz=200*(-1/x^2)*dx
here dx=32
At x=100, sec z=AC/BC=223.61/100=2.24.
Substituting the values in the above equation.
dz= -200*32/(224^2)= -0.127
Change in angle of depression at 100ft = -0.127
A______________________
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B
Here A is the position of the woman. AB is the cliff of height 200 ft. AB=200. C is the location of the boat at any time. BC=x increases as the boat moves away from the cliff at 32 ft/s. The angle of depression 'z' is the angle made by AC with the horizontal at A.
By alternate angles theorem angle ACB=z
tan z=AB/BC=200/x
differentiating both sides partially,
sec^2(z)*dz=200*(-1/x^2)*dx
here dx=32
At x=100, sec z=AC/BC=223.61/100=2.24.
Substituting the values in the above equation.
dz= -200*32/(224^2)= -0.127
Change in angle of depression at 100ft = -0.127