You are here:

Calculus/Differential Calculus


The base of an isosceles triangle is 8 feet long. If the altitude is 6 feet long and is increasing 3 inches per minute, at what rate are the base angles changing?

Hi Christine,
The first thing we have to do is relate the base angle to the altitude. Consider half of the isosceles triangle to one side and you have a right-angled triangle with a 4 feet long base. If the base angle is θ and the altitude H, then the relationship between them is given by;
tanθ = H/4
To find the rates of change we differentiate both sides with respect to time t.
d(tanθ)/dt = d(H/4)/dt
d(tanθ)/dθ  .  dθ/dt = (1/4)dH/dt
secēθ dθ/dt = (1/4)dH/dt
And we immediately see that the rate of change of the base angle dθ/dt depends on the value of dH/dt and secēθ at that time. So, at H = 6 feet
tanθ = 6/4 = 1.5
We can use this to find secēθ from the trigonometric identity
secēθ = 1 + tanēθ
= 1 + 2.25
= 3.25
Therefore, at the moment when the altitude is 6 feet and increasing at 3 inches/minute
dH/dt = 3 inches/minute = 0.25 feet/minute
3.25 dθ/dt = (1/4)(0.25)
dθ/dt = (0.25/4)/3.25
= 0.019 radians/minute

Be careful not to write the unit for the angle in degrees since d(tanθ)/dθ is only secēθ when the angle unit is in radians.



All Answers

Answers by Expert:

Ask Experts


Ahmed Salami


I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.


Aspiring theoretical physicist. I have been doing maths and physics all my life.

I teach mathematics and engineering physics.

©2016 All rights reserved.