Asked by Crain
Officer Weihe is in a patrol car sitting 25ft from the highway and observes Jacob in his car approaching. At a particular instant, t seconds, Jacob is x feet down the highway. The line of sight to Jacob makes an angle of theta radians to a perpendicular to the highway.
Find (dtheta/dx). Use the chain rule to write an equation for (dtheta/dt).
When Jacob is at x=100ft, the angle is ovserved to be changing at a rate (dtheta/dt)=-.05rad/sec. How fast is he going?
Find (dtheta/dx). Use the chain rule to write an equation for (dtheta/dt).
When Jacob is at x=100ft, the angle is ovserved to be changing at a rate (dtheta/dt)=-.05rad/sec. How fast is he going?
Answers
Answered by
Crain
I tried and found (dtheta/dx) but could not find the rest.
Answered by
Reiny
I assume you made a diagram showing the highway and the road at 90° and the cop at 225 ft from the highway.
let the moving car's position be x ft from the intersection, and Ø be the angle.
then x/225 = tanØ
x = 225tanØ
1 = 225 sec^2Ø dØ/dx
dØ/dx = 1/sec^2 Ø
dx/dt = 225sec^2 Ø dØ/dt
when x = 100
tanØ = 100/225 = 4/9
then in my triangle the hypotenuse is √97
cosØ = 9/√97
cos^2 Ø = 81/97
sec^2 Ø = 97/81 and dØ/dt = -.05
then:
dx/dt = 225(97/81)(-.05) = -485/36
= appr -13.47 ft/sec
so the car is approaching the intersection at 13.47 ft/sec
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let the moving car's position be x ft from the intersection, and Ø be the angle.
then x/225 = tanØ
x = 225tanØ
1 = 225 sec^2Ø dØ/dx
dØ/dx = 1/sec^2 Ø
dx/dt = 225sec^2 Ø dØ/dt
when x = 100
tanØ = 100/225 = 4/9
then in my triangle the hypotenuse is √97
cosØ = 9/√97
cos^2 Ø = 81/97
sec^2 Ø = 97/81 and dØ/dt = -.05
then:
dx/dt = 225(97/81)(-.05) = -485/36
= appr -13.47 ft/sec
so the car is approaching the intersection at 13.47 ft/sec
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