A road perpendicular to a highway leads to a farmhouse located 6 mile away. An automobile traveling on the highway passes through this intersection at a speed of 55mph.
How fast is the distance between the automobile and the farmhouse increasing when the automobile is 2 miles past the intersection of the highway and the road?
I don't even know where to begin, can someone PLEASE PLEASE PLEASE explain this step by step!
Thank you!!!!
as usual, begin by drawing a diagram. Clearly, the distance z is the hypotenuse of a triangle with sides 2 and 6.
Now, at time t hours from when the car passed the intersection, the distance z is
z^2 = (55t)^2 + 6^2
so, when the car is 2 miles past the intersection, t=2/55, and z=√40
2z dz/dt = 2(55t)(55)
2√40 dz/dt = 2(55 * 2/55)(55)
dz/dt = 17.38 mi/hr
To solve this problem, we can apply the concept of related rates. The idea is to express the given quantities and the desired rate of change in terms of a common variable, and then differentiate the equation.
Let's denote the distance between the automobile and the farmhouse as "x" and let "t" represent time. We know that the speed of the automobile, which we can denote as "dx/dt," is 55 mph.
From the information given, we can draw a right triangle with one side representing the distance between the farmhouse and the intersection (which we'll call "y") and another side representing the distance between the automobile and the intersection (which we'll call "z"). The hypotenuse of this right triangle can then be expressed as the sum of y and z.
Using the Pythagorean theorem, we have y^2 + z^2 = x^2.
Now, to differentiate this equation with respect to time (t), we have:
2y(dy/dt) + 2z(dz/dt) = 2x(dx/dt).
Since we are looking for the rate of change of the distance between the automobile and the farmhouse, which is dx/dt, and we want it when the automobile is 2 miles past the intersection, we can substitute the given values into the equation.
From the triangle, we have y = 6 miles, z = 2 miles, and x represents the distance between the automobile and the farmhouse. We also know that dx/dt = 55 mph.
Substituting these values into the equation, we get:
2(6)(0) + 2(2)(dz/dt) = 2x(55).
Simplifying the equation, we have:
4(dz/dt) = 110x.
Now, we need to determine the value of x (the distance between the automobile and the farmhouse) when the automobile is 2 miles past the intersection. Since the automobile passed the intersection, x at this moment would be 6 miles + 2 miles = 8 miles.
Substituting this value into the equation, we have:
4(dz/dt) = 110(8).
Simplifying further, we get:
4(dz/dt) = 880.
Finally, we solve for dz/dt to find the rate at which the distance between the automobile and the farmhouse is changing:
dz/dt = 880/4 = 220 mph.
Therefore, when the automobile is 2 miles past the intersection, the distance between the automobile and the farmhouse is increasing at a rate of 220 mph.