Angular Rate of Change (No Calculator)

Question

A squirrel is climbing up a tree attempting to escape from a dog that is sitting 5 feet from the base of the tree. The position of the squirrel at any time t > 0 is represented by the equation h ( t ) = t 2 + 3. At the moment the squirrel is 5 feet from the ground, how fast is the angle of elevation from the dog to the squirrel changing?

Answer 1

construct a right triangle with sides d=5 feet for the base

h=height = 5 feet

z = distance from the dog to the squirrel

h(t) = t^2 +3 = 5,

t^2 =5-3=2

t =sqr2 seconds = about 1.414 when the squirrel is 5 feet up

d= 5 feet = distance from the dog to the tree, directly below the squirrel

it's an isosceles right triangle, 45-45-90 degree triangle

the angle = 45 degrees = pi/4 radians at time t=2sqr2 seconds

rate of change of h = 2t = 2(2sq2) = 4sqr2

z= sqr(5^2 +5^2) = sqr50 = 5sqr2

use the Pythagorean Theorem

z^2 = h^2 +d^2

take the derivative with respect to time, and divide by 2

zz' = hh' +dd'

plug in the known values

5sqr2(z') = 5(4sqr2) + 5(0)= 20sqr2

z' = 4 feet per second

angle = arccos^-1(h/z) = arccos^-1(5/5sqr2) = arccos^-1(1/sqr2) = arccos^-1(sqr2)/2)=about arccos^-1(.707) =45 degrees

= pi/4

= about 3.141593/4 radians=about 0.785398 radians

= about 0.785 radians

or, angle = arctan^-1(h/d) = arctan^-1(5/5) = arctan^-1(1) = 45 degrees = pi/4 = about .79 radians

one second later the height will be (sqr2 +1)^2 + 3 = 3+2sqr2+3 = 6+2sqr2

= about 6+2.828 = 8.828 feet high

it moved 8.828-5 = 3.828 feet in one second

then the angle's tangent = 8.828/5

the angle = arctan(1.7656)= about 60.474 degrees

60.474-45 = 15.474 degrees per 1 second

= about .27 radians per second

one second earlier the height would be (sqr2-1)^3 +3= 2-2sqr2+1+3 =6-2sqr2 = about 6-2.828= 3.172

then the angle = tan^-1(3.172/5) = tan^-1(.6344) = about 32.391

the angle increased 45-32.391= 12.609 degrees =.22 radians

average over 2 second= .(.22+,27)/2 = .245= about .25 radians per second

or (15.5+12.6)/2 =about 14.1 degrees per second

the angle's instantaneous rate of change when h=5, should be between .22 and .27 radians per second

and probably closer to the average, this is a check on whatever solution you may get doing derivatives of inverse trig function.

such as

sqr2/5 = about 1.414/5

= 0.28 radians per second

= about 16.2 degrees per second

Answer 2

h(t) = h = t² + 3 = height of the squirrel at time t

When h = 5, t² + 3 = 5. So, t = √2. Therefore, dh/dt = 2t = 2√2.

Let θ be the angle formed by the line segment from the dog to the base of the tree and the line segment from the dog to the squirrel at time t.

tanθ = h/5 When h = 5, tanθ = 5/5 = 1. So, θ = π/4

sec²θ (dθ/dt) = (1/5)(dh/dt)

2(dθ/dt) = (2√2)/5

dθ/dt = √2/5