if the car has gone x miles, the distance z to the farmhouse is
z^2 = x^2 + 36
so, find z(10) and then
z dz/dt = x dx/dt
and plug in your numbers.
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 75mph.
How fast is the distance between the automobile and the farmhouse increasing when the automobile is 10
miles past the intersection of the highway and the road?
2 answers
h = highway distance from intersection
d = auto-to-farmhouse distance
by Pythagoras ... d^2 = h^2 + 6^2
differentiating implicitly
... 2 d dd/dt = 2 h dh/dt
dd/dt = h / d * dh/dt
d = auto-to-farmhouse distance
by Pythagoras ... d^2 = h^2 + 6^2
differentiating implicitly
... 2 d dd/dt = 2 h dh/dt
dd/dt = h / d * dh/dt
2 answers