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?
I tried and found (dtheta/dx) but could not find the rest.
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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To find (dθ/dx), we need to use trigonometry. Let's construct a right triangle to represent the situation:
* Jacob's car (x ft)
/|
/ |
/ |
Height (25 ft) / |
/ θ |
/ |
Officer Weihe /______|
x ft
In this triangle, we know that the hypotenuse is x ft (distance along the highway) and the opposite side is 25 ft (distance from Officer Weihe's car to the highway). Therefore, we can use the tangent function to relate the angle θ to x:
tan(θ) = 25/x
To find (dθ/dx), we will differentiate this equation implicitly with respect to x:
sec^2(θ) * (dθ/dx) = -25/x^2
Now, let's write an equation for (dθ/dt) using the chain rule. We can relate θ to t by noting that Jacob's position x is a function of t:
x = f(t)
Differentiating both sides of this equation with respect to t, we have:
(dx/dt) = (df/dt)
Substituting the chain rule into our previous equation (sec^2(θ) * (dθ/dx) = -25/x^2), we can express (dθ/dt) in terms of (dx/dt) and (dθ/dx):
sec^2(θ) * (dθ/dx) * (dx/dt) = -25/x^2
Now, let's solve for (dx/dt) when x = 100 ft and (dθ/dt) = -0.05 rad/sec. We can assume that θ is small, so tan(θ) ≈ θ (in radians):
tan(θ) ≈ θ
25/x ≈ θ
θ ≈ 25/x
We can substitute this approximation for θ in our earlier equation:
sec^2(θ) * (dθ/dx) * (dx/dt) = -25/x^2
Let's plug in the given values:
sec^2(25/100) * (dθ/dx) * (dx/dt) = -25/100^2
The secant squared of 25/100 is 1.00374187397 (approximately). Simplifying the equation further:
1.00374187397 * (dθ/dx) * (dx/dt) = -25/100^2
(dθ/dx) * (dx/dt) = -0.000625
We want to find the value of (dx/dt). Dividing both sides of the equation by (dθ/dx), we get:
(dx/dt) = -0.000625 / (dθ/dx)
So, to find how fast Jacob is going, we need to know (dθ/dx). However, the expression for (dθ/dx) depends on the specific values of x and θ.