Related rates waliking away from light towards building

[dethnode]

Homework Statement

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?

Homework Equations

using relative triangles

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?

 

[lanedance]

wasn't this in the other posting?

 

[dethnode]

yes i had already posted this, i have a new posting out titled related rates kite

 

[HallsofIvy]

Homework Statement

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?

Homework Equations

using relative triangles

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?

Yes, that is correct.

 

[HallsofIvy]

yes i had already posted this, i have a new posting out titled related rates kite

Please don't post the same thing more than once!