Tuesday, October 10, 2017

Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 19

How fast is the base of the triangle changing when the altitude is 10cm and the area is 100cm2

Illustration







Given:

dhdt=1cm/min

dAdt=2cm2/min

Required: dbdt when h=10cm and A=100cm2

Solution:


A=12bh,area of triangledAdt=12[bdhdt+hdbdt](Derivative with respect to time)dbdt=2dAdtbdhdth


To get the value of b, we will use the formula of area of triangle


A=12bhb=2Ah=2(100)10=20cm


To solve for the unknown,


dbdt=2(2)20(1)10dbdt=1.6cm/min(It means that the rate is decreasing)


The length of the shadow is decreasing at a rate of 0.6m/s when the man is 4m from the building.

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