dethnode
Oct13-09, 03:29 AM
1. The problem statement, all variables and given/known data
a spotlight won the ground shines on a wall 12 m away if am man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s how fast is the length of his shadow on the building decreasing when he si 4 m from the building?
2. Relevant equations
using relative triangles
3. The attempt at a solution
trying to learn related rates as well, this is what i got tell me if i am wrong here...
draw triangle ABC with A being the light, B being the base of building, and C being top of shaddow/building.
the second triangle is formend with the man and the light, using ADE, D being the mans feet and E being his head at 2 m height.
using the 2 meter horizontal from the mans height we have two relative triangles.
call the range from the light (line AD) x and call the building/shaddow (line BC) y
using the two triangles we can infer that 2/x=y/12 or xy=24
if we then differentiate relative to time 0=dx/dt * y + dy/dt * x
we then plug in 1.6 for dx/dt and 8 for x(range from the light not the building); and y=3 (using xy=24 @ x=8) and solve for dy/dt= -.6m/s which should be negative.
Is this correct?
a spotlight won the ground shines on a wall 12 m away if am man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s how fast is the length of his shadow on the building decreasing when he si 4 m from the building?
2. Relevant equations
using relative triangles
3. The attempt at a solution
trying to learn related rates as well, this is what i got tell me if i am wrong here...
draw triangle ABC with A being the light, B being the base of building, and C being top of shaddow/building.
the second triangle is formend with the man and the light, using ADE, D being the mans feet and E being his head at 2 m height.
using the 2 meter horizontal from the mans height we have two relative triangles.
call the range from the light (line AD) x and call the building/shaddow (line BC) y
using the two triangles we can infer that 2/x=y/12 or xy=24
if we then differentiate relative to time 0=dx/dt * y + dy/dt * x
we then plug in 1.6 for dx/dt and 8 for x(range from the light not the building); and y=3 (using xy=24 @ x=8) and solve for dy/dt= -.6m/s which should be negative.
Is this correct?