Applications of differentiation

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SUMMARY

The discussion centers on solving a problem involving the application of differentiation to the position function \( x = \cos(\theta) + \sqrt{16 - \sin^2(\theta)} \). The user seeks assistance in finding the velocity by computing the first derivative and applying the chain rule. The derivative \( \frac{df}{\theta} = -\sin(\theta)\left(1 + \frac{\cos(\theta)}{\sqrt{16 - \sin^2(\theta)}}\right) \) has been correctly derived, but the user is unsure how to proceed with the given \( \theta = \frac{\pi}{2}t \) to find \( \frac{d\theta}{dt} \).

PREREQUISITES
  • Understanding of differentiation and derivatives
  • Familiarity with the chain rule in calculus
  • Knowledge of trigonometric functions and their derivatives
  • Ability to interpret parametric equations
NEXT STEPS
  • Calculate \( \frac{d\theta}{dt} \) given \( \theta = \frac{\pi}{2}t \)
  • Apply the chain rule to find the velocity function \( \frac{dx}{dt} \)
  • Set the velocity function equal to zero to find critical points
  • Explore applications of differentiation in physics, particularly in motion analysis
USEFUL FOR

Students and educators in calculus, particularly those focusing on applications of differentiation in physics and motion analysis.

Colin2
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I need help solving all three parts to this question, never seen a question regarding applications of differentiation that is this hard before!

View attachment 4094

All help is much appreciated.
 

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We are given the function:

$$x=\cos(\theta)+\sqrt{16-\sin^2(\theta)}$$

This is the position function...how can we find the velocity so that we may equate it to zero and solve?
 
View attachment 4096

I found the first derivative to obtain the velocity, but I'm stuck here.
I also don't know what to do with the value that is given for theta.
 

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You are neglecting to apply the chain rule. You are given:

$$x=f(\theta)$$

And so we must have:

$$\d{x}{t}=\d{f}{\theta}\cdot\d{\theta}{t}$$

Now, you have correctly computed:

$$\d{f}{\theta}=-\sin(\theta)\left(1+\frac{\cos(\theta)}{\sqrt{16-\sin^2(\theta)}}\right)$$

We are given:

$$\theta=\frac{\pi}{2}t$$

So, what is $$\d{\theta}{t}$$?
 
I kinda forgot that I posted this, anyone still willing to help me solve this just for the sake of solving it?
Any help is much appreciated!
 
Colin said:
I kinda forgot that I posted this, anyone still willing to help me solve this just for the sake of solving it?
Any help is much appreciated!

Hi Colin,

Have you tried to follow what Mark has shown above. Did you find $\frac{d\theta}{dt}$?
 

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