Constant Magnitude
Since
\[|\mathbf a|^2=\mathbf a\cdot\mathbf a,\]
differentiation gives
\[\frac{d}{dt}|\mathbf a|^2=2\mathbf a\cdot\frac{d\mathbf a}{dt}.\]
Therefore, constant magnitude is connected with a zero dot product.
Constant Magnitude and Direction
Necessary and Sufficient Conditions for Vector Functions
Constant magnitude and direction are developed through three IOE past-paper proofs. Separate the necessary and sufficient conditions and complete both directions of each result.
UnitVector Calculus
TopicVector Functions
Question Type4 Mark Problems
Theory Required for These Problems
Constant Direction
For \(\mathbf a\ne\mathbf0\), the unit vector in its direction is
\[\widehat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}.\]
The direction is constant when
\[\frac{d\widehat{\mathbf a}}{dt}=\mathbf0.\]
Parallel Vector Test
If
\[\mathbf a\times\frac{d\mathbf a}{dt}=\mathbf0,\]
then \(d\mathbf a/dt\) is parallel to \(\mathbf a\). Hence, only the magnitude changes and the direction remains fixed.
Necessary and Sufficient
Prove both directions separately:
- Assume the vector has the stated property and obtain the condition.
- Assume the condition and obtain the stated property.
Note: Write both parts. One direction alone is not a complete proof.
Method used: For magnitude, differentiate \(\mathbf a\cdot\mathbf a\). For direction, write \(\mathbf a=|\mathbf a|\widehat{\mathbf a}\) and prove the necessary and sufficient conditions separately.
Continue through the IOE Engineering Mathematics II Digital Workbook. This constant magnitude and direction page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
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4 Marks
Condition for Constant Magnitude
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 9 • 4 Marks
Prove that the necessary and sufficient condition for a vector function \(\mathbf a\) of a scalar variable \(t\) to have constant magnitude is
\[\mathbf a\cdot\frac{d\mathbf a}{dt}=0.\]
Solution
Necessary and sufficient proofResult #1
If \(\vec a\) is a vector function of \(t\), then
\[\frac{d}{dt}(\vec a\cdot\vec a)=2\left[\vec a\cdot\frac{d\vec a}{dt}\right].\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=2\left[\vec a\cdot\frac{d\vec a}{dt}\right].\]
Proof of Result #1
\[\frac{d}{dt}(\vec a\cdot\vec a)=\vec a\cdot\frac{d\vec a}{dt}+\frac{d\vec a}{dt}\cdot\vec a.\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=\vec a\cdot\frac{d\vec a}{dt}+\frac{d\vec a}{dt}\cdot\vec a.\]
Since the dot product is commutative,
\[\frac{d}{dt}(\vec a\cdot\vec a)=\vec a\cdot\frac{d\vec a}{dt}+\vec a\cdot\frac{d\vec a}{dt}.\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=\vec a\cdot\frac{d\vec a}{dt}+\vec a\cdot\frac{d\vec a}{dt}.\]
Therefore,
\[\frac{d}{dt}(\vec a\cdot\vec a)=2\left[\vec a\cdot\frac{d\vec a}{dt}\right].\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=2\left[\vec a\cdot\frac{d\vec a}{dt}\right].\]
We have to prove that
\[\vec a\ \mathrm{has\ constant\ magnitude}\quad\Longleftrightarrow\quad\vec a\cdot\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ constant\ magnitude}\]\[\Longleftrightarrow\ \vec a\cdot\frac{d\vec a}{dt}=0.\]
Necessary condition
Assume that the vector function \(\vec a\) has constant magnitude. Therefore, \(|\vec a|\) is a constant.
Hence,
\[|\vec a|^2=c,\qquad c\ \mathrm{is\ a\ constant}.\]Since \(\vec a\cdot\vec a=|\vec a|^2\),
\[\vec a\cdot\vec a=c.\]Differentiating with respect to \(t\),
\[\frac{d}{dt}(\vec a\cdot\vec a)=\frac{dc}{dt}.\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=\frac{dc}{dt}.\]
By Result #1,
\[2\left[\vec a\cdot\frac{d\vec a}{dt}\right]=0.\]Therefore,
\[\vec a\cdot\frac{d\vec a}{dt}=0.\]Sufficient condition
Assume that
\[\vec a\cdot\frac{d\vec a}{dt}=0.\]We have
\[\vec a\cdot\vec a=|\vec a|^2.\]Differentiating both sides with respect to \(t\),
\[\frac{d}{dt}(\vec a\cdot\vec a)=\frac{d}{dt}|\vec a|^2.\]
\[\frac{d}{dt}(\vec a\cdot\vec a)\]\[=\frac{d}{dt}|\vec a|^2.\]
By Result #1,
\[2\left[\vec a\cdot\frac{d\vec a}{dt}\right]=2|\vec a|\frac{d|\vec a|}{dt}.\]
\[2\left[\vec a\cdot\frac{d\vec a}{dt}\right]\]\[=2|\vec a|\frac{d|\vec a|}{dt}.\]
Therefore,
\[|\vec a|\frac{d|\vec a|}{dt}=\vec a\cdot\frac{d\vec a}{dt}.\]By the assumption in the sufficient condition,
\[|\vec a|\frac{d|\vec a|}{dt}=0.\]Therefore,
\[|\vec a|=0\qquad\mathrm{or}\qquad\frac{d|\vec a|}{dt}=0.\]
\[|\vec a|=0\]\[\mathrm{or}\]\[\frac{d|\vec a|}{dt}=0.\]
If \(|\vec a|=0\), then \(\vec a=0\), the zero vector, which is the trivial case. Hence, for a non-zero vector function,
\[\frac{d|\vec a|}{dt}=0.\]Therefore,
\[|\vec a|\ \mathrm{is\ a\ constant}.\]i.e., \(\vec a\) has constant magnitude.
\[\vec a\ \mathrm{has\ constant\ magnitude}\quad\Longleftrightarrow\quad\vec a\cdot\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ constant\ magnitude}\]\[\Longleftrightarrow\ \vec a\cdot\frac{d\vec a}{dt}=0.\]
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4 Marks
Condition for Constant Direction
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 9 • 4 Marks
Prove that the necessary and sufficient condition for a vector function \(\mathbf a\) of a scalar variable \(t\) to have constant direction is
\[\mathbf a\times\frac{d\mathbf a}{dt}=0.\]
Solution
Necessary and sufficient proofResult #2
If \(\vec a\) is a vector function of \(t\), then
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
Proof of Result #2
The unit vector along \(\vec a\) is
\[\hat a=\frac{\vec a}{|\vec a|}.\]Therefore,
\[\vec a=|\vec a|\hat a.\tag{1}\]Here, \(|\vec a|\) is a scalar function and \(\hat a\) is a vector function.
For a scalar function \(u\) and a vector function \(\vec v\),
\[\frac{d}{dt}(u\vec v)=u\frac{d\vec v}{dt}+\frac{du}{dt}\vec v.\]
\[\frac{d}{dt}(u\vec v)\]\[=u\frac{d\vec v}{dt}+\frac{du}{dt}\vec v.\]
Differentiating (1),
\[\frac{d\vec a}{dt}=\frac{d}{dt}\left(|\vec a|\hat a\right).\]
\[\frac{d\vec a}{dt}\]\[=\frac{d}{dt}\left(|\vec a|\hat a\right).\]
Using the product rule,
\[\frac{d\vec a}{dt}=|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a.\]
\[\frac{d\vec a}{dt}\]\[=|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a.\]
Taking the cross product with \(\vec a\),
\[\vec a\times\frac{d\vec a}{dt}=\vec a\times\left(|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=\vec a\times\left(|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a\right).\]
Distributing the cross product,
\[\vec a\times\frac{d\vec a}{dt}=\vec a\times|\vec a|\frac{d\hat a}{dt}+\vec a\times\frac{d|\vec a|}{dt}\hat a.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=\vec a\times|\vec a|\frac{d\hat a}{dt}+\vec a\times\frac{d|\vec a|}{dt}\hat a.\]
Substituting \(\vec a=|\vec a|\hat a\) from (1),
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|\hat a\times|\vec a|\frac{d\hat a}{dt}+|\vec a|\hat a\times\frac{d|\vec a|}{dt}\hat a.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|\hat a\times|\vec a|\frac{d\hat a}{dt}\]\[{}+|\vec a|\hat a\times\frac{d|\vec a|}{dt}\hat a.\]
Taking the scalar factors outside,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+|\vec a|\frac{d|\vec a|}{dt}(\hat a\times\hat a).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)\]\[{}+|\vec a|\frac{d|\vec a|}{dt}(\hat a\times\hat a).\]
Since \(\hat a\times\hat a=0\),
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+0.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+0.\]
Therefore,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
We have to prove that
\[\vec a\ \mathrm{has\ fixed\ direction}\quad\Longleftrightarrow\quad\vec a\times\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ fixed\ direction}\]\[\Longleftrightarrow\ \vec a\times\frac{d\vec a}{dt}=0.\]
Necessary condition
Assume that the vector function \(\vec a\) has fixed direction.
Therefore, \(\hat a\) has fixed direction.
Also,
\[|\hat a|=1,\]i.e., \(\hat a\) has constant magnitude.
Therefore, \(\hat a\) has constant magnitude and fixed direction.
Hence, \(\hat a\) is a constant vector.
Therefore,
\[\frac{d\hat a}{dt}=0.\tag{1}\]By Result #2,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
Using (1),
\[\vec a\times\frac{d\vec a}{dt}=0.\]Sufficient condition
Assume that
\[\vec a\times\frac{d\vec a}{dt}=0.\]By Result #2,
\[|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)=0.\]Therefore,
\[|\vec a|^2=0\qquad\mathrm{or}\qquad\hat a\times\frac{d\hat a}{dt}=0.\]
\[|\vec a|^2=0\]\[\mathrm{or}\]\[\hat a\times\frac{d\hat a}{dt}=0.\]
If \(|\vec a|^2=0\), then \(\vec a=0\), the zero vector, which is the trivial case. Hence, for a non-zero vector function,
\[\hat a\times\frac{d\hat a}{dt}=0.\tag{2}\]Now,
\[|\hat a|=1.\]i.e., \(\hat a\) has constant magnitude.
By Theorem #1,
\[\hat a\cdot\frac{d\hat a}{dt}=0.\tag{3}\]From (2), \(d\hat a/dt\) is parallel to \(\hat a\).
From (3), \(d\hat a/dt\) is perpendicular to \(\hat a\).
Equations (2) and (3) are possible simultaneously only if
\[\frac{d\hat a}{dt}=0.\]Therefore, \(\hat a\) is a constant vector.
Hence, \(\hat a\) has fixed direction.
Therefore, \(\vec a\) has fixed direction.
\[\vec a\ \mathrm{has\ fixed\ direction}\quad\Longleftrightarrow\quad\vec a\times\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ fixed\ direction}\]\[\Longleftrightarrow\ \vec a\times\frac{d\vec a}{dt}=0.\]
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4 Marks
Condition for Constant Direction
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 9 • 4 Marks
Prove that the necessary and sufficient condition for a vector function \(\mathbf a\) of a scalar variable \(t\) to have constant direction is
\[\mathbf a\times\frac{d\mathbf a}{dt}=0.\]
Solution
Necessary and sufficient proofResult #2
If \(\vec a\) is a vector function of \(t\), then
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
Proof of Result #2
The unit vector along \(\vec a\) is
\[\hat a=\frac{\vec a}{|\vec a|}.\]Therefore,
\[\vec a=|\vec a|\hat a.\tag{1}\]Here, \(|\vec a|\) is a scalar function and \(\hat a\) is a vector function.
For a scalar function \(u\) and a vector function \(\vec v\),
\[\frac{d}{dt}(u\vec v)=u\frac{d\vec v}{dt}+\frac{du}{dt}\vec v.\]
\[\frac{d}{dt}(u\vec v)\]\[=u\frac{d\vec v}{dt}+\frac{du}{dt}\vec v.\]
Differentiating (1),
\[\frac{d\vec a}{dt}=\frac{d}{dt}\left(|\vec a|\hat a\right).\]
\[\frac{d\vec a}{dt}\]\[=\frac{d}{dt}\left(|\vec a|\hat a\right).\]
Using the product rule,
\[\frac{d\vec a}{dt}=|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a.\]
\[\frac{d\vec a}{dt}\]\[=|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a.\]
Taking the cross product with \(\vec a\),
\[\vec a\times\frac{d\vec a}{dt}=\vec a\times\left(|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=\vec a\times\left(|\vec a|\frac{d\hat a}{dt}+\frac{d|\vec a|}{dt}\hat a\right).\]
Distributing the cross product,
\[\vec a\times\frac{d\vec a}{dt}=\vec a\times|\vec a|\frac{d\hat a}{dt}+\vec a\times\frac{d|\vec a|}{dt}\hat a.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=\vec a\times|\vec a|\frac{d\hat a}{dt}+\vec a\times\frac{d|\vec a|}{dt}\hat a.\]
Substituting \(\vec a=|\vec a|\hat a\) from (1),
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|\hat a\times|\vec a|\frac{d\hat a}{dt}+|\vec a|\hat a\times\frac{d|\vec a|}{dt}\hat a.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|\hat a\times|\vec a|\frac{d\hat a}{dt}\]\[{}+|\vec a|\hat a\times\frac{d|\vec a|}{dt}\hat a.\]
Taking the scalar factors outside,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+|\vec a|\frac{d|\vec a|}{dt}(\hat a\times\hat a).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)\]\[{}+|\vec a|\frac{d|\vec a|}{dt}(\hat a\times\hat a).\]
Since \(\hat a\times\hat a=0\),
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+0.\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)+0.\]
Therefore,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
We have to prove that
\[\vec a\ \mathrm{has\ fixed\ direction}\quad\Longleftrightarrow\quad\vec a\times\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ fixed\ direction}\]\[\Longleftrightarrow\ \vec a\times\frac{d\vec a}{dt}=0.\]
Necessary condition
Assume that the vector function \(\vec a\) has fixed direction.
Therefore, \(\hat a\) has fixed direction.
Also,
\[|\hat a|=1,\]i.e., \(\hat a\) has constant magnitude.
Therefore, \(\hat a\) has constant magnitude and fixed direction.
Hence, \(\hat a\) is a constant vector.
Therefore,
\[\frac{d\hat a}{dt}=0.\tag{1}\]By Result #2,
\[\vec a\times\frac{d\vec a}{dt}=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
\[\vec a\times\frac{d\vec a}{dt}\]\[=|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right).\]
Using (1),
\[\vec a\times\frac{d\vec a}{dt}=0.\]Sufficient condition
Assume that
\[\vec a\times\frac{d\vec a}{dt}=0.\]By Result #2,
\[|\vec a|^2\left(\hat a\times\frac{d\hat a}{dt}\right)=0.\]Therefore,
\[|\vec a|^2=0\qquad\mathrm{or}\qquad\hat a\times\frac{d\hat a}{dt}=0.\]
\[|\vec a|^2=0\]\[\mathrm{or}\]\[\hat a\times\frac{d\hat a}{dt}=0.\]
If \(|\vec a|^2=0\), then \(\vec a=0\), the zero vector, which is the trivial case. Hence, for a non-zero vector function,
\[\hat a\times\frac{d\hat a}{dt}=0.\tag{2}\]Now,
\[|\hat a|=1.\]i.e., \(\hat a\) has constant magnitude.
By Theorem #1,
\[\hat a\cdot\frac{d\hat a}{dt}=0.\tag{3}\]From (2), \(d\hat a/dt\) is parallel to \(\hat a\).
From (3), \(d\hat a/dt\) is perpendicular to \(\hat a\).
Equations (2) and (3) are possible simultaneously only if
\[\frac{d\hat a}{dt}=0.\]Therefore, \(\hat a\) is a constant vector.
Hence, \(\hat a\) has fixed direction.
Therefore, \(\vec a\) has fixed direction.
\[\vec a\ \mathrm{has\ fixed\ direction}\quad\Longleftrightarrow\quad\vec a\times\frac{d\vec a}{dt}=0.\]
\[\vec a\ \mathrm{has\ fixed\ direction}\]\[\Longleftrightarrow\ \vec a\times\frac{d\vec a}{dt}=0.\]