Object passes directly above radar at an altitude 6mile. if object speed is 500mph(only using differentiation)

Question

(only using differentiation)
1) how fast is distance changing after 1/2hr (ans. 500mph)
2) how fast is distance between object and radar changing when object passes directly above the radar. (ans. 0)
3) suppose that line through the radar and object make an angle Θ with horizontal. how fast is Θ changing 10min after object passes over radar. (need help with this one)

give hints and ans for 3 part

Answer 1

Well 10 mins after passing over head the object would have travelled 83.333 miles (1/6 of an hour) and the TAN of the angle Θ is 6/83.333 at that point. Already the angle is pretty damned small but one minute later the TAN of Θ would be 6/91.66743 or thereabouts.

Answer 2

trig will work here
1. a right triangle has 6m for a , 250m for b and sq rt (36+250^2) for c
so change in (c) (distance ) relative to its horz speed (b) will be equal to the derivative of fx where y=fx
y is c
x is b
with a as a constant
so if
c= sqrt ( a^2 + b^2)
or fx is
y= sqrt( 36+x^2) the dy/dx of this ....
is dy/dx= 1/2 ( 2x )/sqrt(36+ x^2)
or
b / sqrt(36+ b^2)
or
250/sqrt(62,536)
or
.999 so change is .999 times 500mph or 499.8mph

for number 2 we intuitively know 90 deg movement doesn't change the distance at all for very small intervals the numbers would be found by taking the dx/dy of cos(x) at 0 witch is equal to -sin of zero wtich is zero so answer to number 2 is zero
for number 3 triangle is a= 6m b= 1/6 hour times 500mph or 83.33m
c= sq rt of (36+ (500/6)^2) or 83.5
a/c = .997
if y= fx and O is y and x is a/b ? then y= inv sin(a/c)
or O= inv sin (.997)= 85.8 degs that's the angle at 1/6 hour
the dy/dx of inv sin(a/c) is not the inv cos (a/c)
and i don't know off hand the dy/dx of the inv sin function
its in the book
find it and use .997 for x and solve for y
and that's the change of O at that time 1/6 hour from overhead

Answer 3

t = elapsed time after object passes over radar

h_t = object's horizontal distance from radar's vertical as a
function of time

theta_t = angle between radar's sightline and horizontal as a
function of time

h_t = 500t; tan (theta_t) = 6/h_t = 6/500t

cot (theta_t) = 1/tan (theta_t) = 500t/6

cot ' (theta_t) = -csc^2 (theta_t) * d(theta_t)/t = 500/6

d(theta_t)/dt = - 500/6 csc^2 (theta_t)

sin^2 (theta_t) = 1/csc^2 (theta_t)


Substitute into expression for d(theta_t)/dt:

d(theta_t)/dt = -500 sin^2 (theta_t)/6


When t = 10 min = 1/6 h, then.....

h_t = 500*(1/6) = 500/6 miles

theta_t = arctan (6/h_t) = arctan (36/500)

~ 4.1182 degrees


d(theta_t)/dt = -500/6 * sin^2 (4.1182)

~ -0.4298 radians/hour or

~ -24.624 degrees/hour

This makes sense because theta is at a max of 90 degrees when the object is directly overhead the radar and decreases as the horizontal distance increases