My Math Forum In related-type-problems, why must the expression be differentiated first...?

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 October 27th, 2014, 04:01 PM #1 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 In related-type-problems, why must the expression be differentiated first...? ...prior to substituting values? Example: A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?
 October 27th, 2014, 04:38 PM #2 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus Because the differentiation is the rate of change of a variable respect to another, usually the time.. For example, the speed is the rate of change of position of an object with respect to time. In other words: $\displaystyle v = {\text{d}x \over \text{d}t}$ Acceleration is defined as the rate of change of speed over time: $\displaystyle a = {\text{d}v \over \text{d}t}$ etc. If I am not wrong, you have to solve a differential equation in your problem. Last edited by szz; October 27th, 2014 at 05:05 PM.
October 27th, 2014, 05:16 PM   #3
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Quote:
 Originally Posted by lm1988 ...prior to substituting values? Example: A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?
A rate is a derivative w/r to time ... the idea is to determine how the rates are related, hence the name "related rates" problems.

Also, substituting constant values in for variables that change leads to derivatives equal to zero, which is not the case. It is ok to substitute constants for quantities that do not change, like the ladder length in your example.

Let x = distance the foot of the ladder is from the wall
$\displaystyle \theta$ = angle formed by the foot of the ladder and the ground.

$\displaystyle \cos{\theta} = \frac{x}{10}$

Take the derivative w/r to time ...

$\displaystyle -\sin{\theta} \cdot \frac{d\theta}{dt}=\frac{1}{10} \cdot \frac{dx}{dt}$

Now perform some right triangle trig to determine the value of $\displaystyle \sin{\theta}$ when x = 6 ft, sub in 1 ft/ sec for $\displaystyle \frac{dx}{dt}$, and solve for $\displaystyle \frac{d\theta}{dt}$.

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