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HERE del(theta)/del(x) means partial derivative of theta with respect to x.Please Help with detailed solution...Thanks in advance...

Note by Raja Metronetizen 3 years, 10 months ago

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Disregard my previous comment, this can be proved. We will begin with (ii), (though it doesn't really matter) because we will use the result of (ii) to prove (i).

\((x)^2 + (y)^2 = r^2\) \(\Rightarrow\) \(\frac{\partial} {\partial x}\)\(((x)^2 + (y)^2)\) = \((2r)(\frac{\partial r} {\partial x})\)

\(\Rightarrow\) \(2x = (2r)(\frac{\partial r} {\partial x})\)

\(\Rightarrow\) \(\frac{x} {r} = \frac {\partial r} {\partial x}\)

Note that \(cos(\theta)=\frac{x} {r}\),

Hence \(cos(\theta)=\frac {\partial r} {\partial x}\)

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac {\partial x} {\partial r} = \frac {\partial (rcos(\theta))} {\partial r}\)

\(\Rightarrow\) \(\frac{\partial x} {\partial r} = \frac {\partial r} {\partial x} = cos(\theta) \)

This proves (ii). Now for (i):

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac {\partial x} {\partial \theta} = \frac {\partial} {\partial \theta} (rcos(\theta))\) = \(-rsin(\theta)\)

\(\Rightarrow\) \((\frac {1} {r})\frac{\partial x} {\partial \theta}= -sin(\theta)\)

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac{\partial x} {\partial x}= \frac{\partial} {\partial x} (rcos(\theta)\)

\(\Rightarrow\) \(1= cos(\theta)\frac{\partial r} {\partial x} -rsin(\theta) \frac {\partial \theta} {\partial x}\)

In (ii) we showed that: \(\frac{\partial r} {\partial x} = cos(\theta)\), hence:

\(1=cos^2(\theta) - rsin(\theta)\frac {\partial \theta} {\partial x}\)

\(\Rightarrow\) \(\frac{1-cos^2(\theta)} {-rsin(\theta)}=\frac{\partial \theta} {\partial x}\)

\(\Rightarrow\) \(\frac {-sin(\theta)} {r} = \frac {\partial \theta} {\partial x}\)

\(\Rightarrow\) \(r\frac{\partial \theta} {\partial x} = -sin(\theta) = \frac {1} {r} \frac {\partial x} {\partial \theta}\)

This proves (i). Q.E.D. – Ethan Robinett · 3 years ago

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TopNewestDisregard my previous comment, this can be proved. We will begin with (ii), (though it doesn't really matter) because we will use the result of (ii) to prove (i).

\((x)^2 + (y)^2 = r^2\) \(\Rightarrow\) \(\frac{\partial} {\partial x}\)\(((x)^2 + (y)^2)\) = \((2r)(\frac{\partial r} {\partial x})\)

\(\Rightarrow\) \(2x = (2r)(\frac{\partial r} {\partial x})\)

\(\Rightarrow\) \(\frac{x} {r} = \frac {\partial r} {\partial x}\)

Note that \(cos(\theta)=\frac{x} {r}\),

Hence \(cos(\theta)=\frac {\partial r} {\partial x}\)

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac {\partial x} {\partial r} = \frac {\partial (rcos(\theta))} {\partial r}\)

\(\Rightarrow\) \(\frac{\partial x} {\partial r} = \frac {\partial r} {\partial x} = cos(\theta) \)

This proves (ii). Now for (i):

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac {\partial x} {\partial \theta} = \frac {\partial} {\partial \theta} (rcos(\theta))\) = \(-rsin(\theta)\)

\(\Rightarrow\) \((\frac {1} {r})\frac{\partial x} {\partial \theta}= -sin(\theta)\)

\(x=rcos(\theta)\) \(\Rightarrow\) \(\frac{\partial x} {\partial x}= \frac{\partial} {\partial x} (rcos(\theta)\)

\(\Rightarrow\) \(1= cos(\theta)\frac{\partial r} {\partial x} -rsin(\theta) \frac {\partial \theta} {\partial x}\)

In (ii) we showed that: \(\frac{\partial r} {\partial x} = cos(\theta)\), hence:

\(1=cos^2(\theta) - rsin(\theta)\frac {\partial \theta} {\partial x}\)

\(\Rightarrow\) \(\frac{1-cos^2(\theta)} {-rsin(\theta)}=\frac{\partial \theta} {\partial x}\)

\(\Rightarrow\) \(\frac {-sin(\theta)} {r} = \frac {\partial \theta} {\partial x}\)

\(\Rightarrow\) \(r\frac{\partial \theta} {\partial x} = -sin(\theta) = \frac {1} {r} \frac {\partial x} {\partial \theta}\)

This proves (i). Q.E.D. – Ethan Robinett · 3 years ago

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