Find the first partial derivative of

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To find the first partial derivatives ∂z/∂x and ∂z/∂y of the equation sin(0x + 5y + z) = 0 at the point (0,0,0), the equation simplifies to 0x + 5y + z = kπ, leading to z = kπ - 5y. The derivatives are calculated as ∂z/∂x = 0 and ∂z/∂y = -5. The significance of k and π arises from the solutions to the sine function being zero at integer multiples of π. The discussion highlights the realization that sin(0) = 0, clarifying the relationship between the variables. Understanding these concepts is crucial for solving similar problems in calculus.
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Homework Statement



Find the first partial derivatives ∂z/∂x and ∂z/∂y of sin(0x+5y+z)=0 at (0,0,0).

Homework Equations



sin(0x+5y+z)=0


The Attempt at a Solution



0x+5y+z=kπ
z=kπ-5y

So,

∂z/∂x= 0 and ∂z/∂y= -5

What I do not understand is WHY 0x+5y+z=kπ is an acceptable equation. I found it from a solution of this online, but with no explanation. What is the significance of k and π, and why can we basically ignore the sin?
 
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You have the relation ##\sin(\theta)=0:\theta=5y+z## right ... so what values of ##\theta## make the relation true?
 
Simon Bridge said:
You have the relation ##\sin(\theta)=0:\theta=5y+z## right ... so what values of ##\theta## make the relation true?

Oh, duh! Sin(0)=0.

Thanks
 
##\sin k\pi=0:k=0,1,2,\cdots##
... sometimes you are staring right at it.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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