The altitude of a triangle is increasing at a rate of

Question

The altitude of a triangle is increasing at a rate of  centimeters/minute while the area of the triangle is increasing at a rate of  square centimeters/minute. At what rate is the base of the triangle changing when the altitude is  centimeters and the area is  square centimeters?

Answer 1

This is a "related rates" problem. Calculus books have a section on it, in the derivatives or differentiation part of the textbook. Maybe a few pages.

Basically, use the Pythagorean Theorem to relate the side lengths of the triangle. hypotenuse squared = sum of squares of the other two sides

Then take the derivative to get 2hh' = 2xx' + 2yy', divide by two to get hh' = xx' + yy'

plug in the values of the two sides and solve for x' = rate of change of the base

Or use Area = A =xy/2, take the derivatives to get A'=(1/2)(xy' + y'x), plug in the values, solve for x'

Area of a triangle = one half base times height

A = xy/2 where y = height=8 A=92, and x = base

92 = x(8)/2 = 4x

x = 92/4 = 23 cm

h^2 = 8^2 + 23^2 = 593

h = sqr593 = about 24.35

h^2 = x^2 + y^2

hh' = xx' + yy'

24.35(h') = 23x' + 8(2.5)

x' = -6.8125

h' = (23(-6.8125)+20)/24.35 = -5.613 = rate of change of the hypotenuse = slightly more than x'

A = (1/2)xy

take the derivative of both sides

A' = (1/2)[xy' + yx') where A'=1.5, y'=2.5

1.5 = (1/2)[23(2.5) + 8x']

3 = 23(2.5) + 8x'

8x' = 3- 57.5 = -54.5

x' = -54.5/8 = -6.8125 cm/minute = rate of change of the base of the triangle

check to see what happens in one minute

y goes from 8 to 8+2.5 = 10.5, A from 92 to 92+1.5 = 93.5,

x goes from 23 to 2(93.5)/10.5 = 17.81

x' = 17.81 - 23 = -5.19 which is close to -6.8125

in less than 4 minutes x goes to zero if x' remained the same, but it doesn't as 92 goes to 92+4(1.5) = 98 in 4 seconds which is impossible if the rates remained the same. x' changes, gets smaller as x gets smaller

or see what happens one minute earlier

y went from 5.5 to 8, A from 90.5 to 92. x goes from 2(90.5)/5.5= 32.9 to 23

x' = 23-32.9= -9.91

x' gets increasingly more negative as A and y get increasingly positive

x' = dx/dt at an instant in time as dt approaches zero

x' = approximately change in x divided by change in t, if the change in t is very small.

Let the change in t = 1 second

then y' = 2.5/60 = .5/12 = 0.041666 and A' = 1.5/60= 0.025

in one second y increases from 8 to 8.04166

and A increases from 92 to 92.025

as x decreases from 23 to 22.887 a decrease of 0.113

0.113 x 60 = 6.8 ft/min. rate of change, virtually the same as at an instant in time as calculating x' in feet per miniute rate of change. 1 second is "close" to an instant in time for the calculations.

Answer 2

Let h be the altitude and b be the base. Let A be the Area.

A = hb/2

Given:

dh/dt = 2.5 cm/min

dA/dt = 1.5 cm²/min

Need db/dt

dA/dt = (hdb/dt + dh/dt b)/2

when h =8 and A = 92

92 = 8b/2 --> b = 23 cm

1.5 = (8 db/dt + 2.5 (23))/2

8db/dt + 57.5 = 3

db/dt = -54.5/8 = -6.8125 cm/min

The base is decreasing by 6.1825 cm/min