Implicit differentiation wrt time

In summary, the distance the water travels horizontally in a lawn sprinkler can be calculated using the formula x=v^2sin(2\theta)/32, where v is the speed of the water and theta ranges from 45 degrees to 135 degrees. The derivative of this function, dx/dt, can be found using the chain rule, resulting in dx/dt=[2v*sin2\theta + 2v^2cos2\theta(d\theta/dt)]/32. This shows that the lawn sprinkler does not water evenly because the velocity of the water changes with the angle of the sprinkler, resulting in different amounts of water being distributed to different parts of the lawn. The part of the lawn that receives
  • #1
rook_b

Homework Statement



"A lawn sprinkler is constructed in such a way that d[itex]\theta[/itex]/dt is constant, where [itex]theta[/itex] ranges between 45 degrees and 135 degrees. The distance the water travels horizontally is

[tex]x=v^2sin(2\theta)/32[/tex]

where v is the speed of the water. Find dx/dt and explain why this lawn sprinkler does not water evenly. What part of the lawn receives the most water?

Homework Equations



The Attempt at a Solution



I took the derivative of the function getting,

[tex]dx/dt=(2v*sin2\theta(dx/dt) +2v^2cos2\theta(dx/dt))32[/tex]

I made a mistake because I can factor out dx/dt, then dividing both sides by dx/dt sets the equation to 1 but I don't want that. Where am I making my mistake? It seems to me that my differentiation was correct but I must have made a mistake. If you want I'll walk through it. This answer isn't the same as the one in the back of the book either, I avoid looking at the back usually and forget the answers are even there.
 
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  • #2
Yes, you did make a mistake. This may help; [tex]\frac{d}{dt}(\sin\theta)=\cos\theta\cdot\frac{d\theta}{dt}[/tex]

Incidentally, isn't [tex]v=\frac{dx}{dt}[/tex]
 
  • #3
Oh! right, v=dx/dt . So now I have [itex]\frac{dx}{dt}=[2v*sin2\theta + 2cos2\theta \frac{d\theta}{dt}v^2]/32^2
[/itex]

Is that right? Because we are taking the derivative of cos wrt theta we must multiply by [itex]\frac{d\theta}{dt}[/itex].
 
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1. What is implicit differentiation wrt time?

Implicit differentiation wrt time is a mathematical technique used to find the derivative of a function with respect to time. It is particularly useful for finding the rate of change of a variable that is not explicitly defined in terms of time.

2. How is implicit differentiation wrt time different from regular differentiation?

Implicit differentiation wrt time differs from regular differentiation in that it involves differentiating both sides of an equation with respect to time. This is done by treating the dependent variable as a function of time, and using the chain rule to differentiate any terms that are not explicitly defined in terms of time.

3. What are some common applications of implicit differentiation wrt time?

Implicit differentiation wrt time is commonly used in physics, engineering, and economics to model and analyze dynamic systems. It is particularly useful for finding the velocity, acceleration, and other rates of change of variables in these systems.

4. Can implicit differentiation wrt time be used for any type of function?

Yes, implicit differentiation wrt time can be used for any type of function, including polynomials, trigonometric functions, and exponential functions. However, it may become more complex for functions with multiple variables or complex expressions.

5. What is the purpose of using implicit differentiation wrt time?

The purpose of using implicit differentiation wrt time is to find the rate of change of a variable that is not explicitly defined in terms of time. This can provide valuable information about how a system is changing over time and can be used to make predictions and optimize processes.

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