Define tangentA tangent line is a straight line that touches a curve exactly at a point on which both (curve and tangent) going in exact same direction. In a curve in space, number of lines may pass touching exactly at a point, so a precise definition of tangent line is required.
DefinitionLet C \( \vec{r}= \vec{r} (t) \) be a space curve.
Also let A and B are two nearby points on C corresponding to the parameter values \(t\) and \(t+ \delta t\) respectively. Then position vector of A and B are respectively given as
Position vector of A = \( \vec{r} (t) \)
Position vector of B = \( \vec{r} (t+ \delta t) \)
Now, slope of secant line AB is
slope of secant line AB = \( \frac{\vec{r}( t+\delta t )-\vec{r}( t )}{\delta t}\)
When, point B is moved toward A then limiting form of secant AB as \(\delta t \to 0 \) , is called tangent to the curve \( C:\vec{r}=\vec{r}( t )\) at A. Hence, slope of tangent line at A is written as:
slope of tangent at A \( =\displaystyle \lim_{\delta t \to 0} \frac{\vec{r}( t+\delta t )-\vec{r}( t )}{\delta t}\)
slope of tangent at A \( =\frac{d \vec{r}( t )}{dt}\)
slope of tangent at A \( =\dot{\vec{r}}( t )\)
So, tangent at A is limiting form of secant line AB as \( B \to A\).
Unit Tangent Vector
Show that \( \vec{t}=\vec{r}' \) is expression of unit tangent vector.
Proof
Let \( C:\vec{r}=\vec{r}(s) \) be a space curve.
Also, let P and Q are two neighboring points on C and their respective position vector be
\( \vec{OP}=\vec{r} \)
and
\( \vec{OQ}=\vec{r}+\delta \vec{r} \)
Then
\( \vec{PQ}=\delta \vec{r} \)
Assume that arc length s of C be measured from suitable base point A, we have
arcAP=s and arcAQ=s+ \( \delta \) s
Then,
arcPQ=\( \delta \) s
Now, tangent at P is defined as
tangent at P= \( \displaystyle \lim_{Q\to P} \vec{PQ} \)
If \( \vec{t} \) is unit tangent vector along \( \vec{PQ} \), then
\( \vec{t}= \displaystyle \lim_{Q \to P} \frac {\vec{PQ}} {|\vec{PQ}|} \)
or
\( \vec{t}= \displaystyle \lim_{Q \to P} \frac {\vec{PQ}} {|\vec{PQ}|} \)
or
\( \vec{t}=\displaystyle \lim_{Q\to P} \frac{\delta \vec{r}}{|\delta \vec{r}|}\)
or
\( \vec{t}=\displaystyle \lim_ {Q\to P} \frac{\delta \vec{r}}{\delta s}.\frac{\delta s}{|\delta \vec{r}|}\)
or
\( \vec{t}=\frac{d\vec{r}}{ds} \times 1\)
or
\( \vec{t}=\vec{r}'\)
Thus, \( \vec{t}=\vec{r}'\) is expression of unit tangent vector.
Examples
-
Find unit tangent vector to a space curve \( \vec{r}=( a \cos t,a \sin t,bt )\).
Solution
Given space curve is
\( \vec{r}=( a \cos t,a \sin t,bt )\)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=( -a \sin t,a \cos t,b )\)
We know that, \( \dot{\vec{r}}\) is tangent vector, so unit tangent vector is
\(\frac{\dot{\vec{r}}}{| \dot{\vec{r}} |}\)
Thus, taking magnitude, we get
\(| \dot{\vec{r}} |=\sqrt{a^2+b^2}\)
Hence, unit vector along tangent to the curve is
\(\vec{t}=\frac{\dot{\vec{r}}}{| \dot{\vec{r}} |}=\frac{1}{\sqrt{a^2+b^2}}( -a \sin t,a \cos t,b )\)
This completes the solution.
Equation of Tangent Line
Deduce equation of tangent on a space curve \(C:\vec{r}=\vec{r}( s )\)
Proof
Let \(C:\vec{r}=\vec{r}( s )\) be a space curve and \(P\) be a point on it with position vector
\( \vec{OP}=\vec{r}\)
Also let \( \vec{t}\) denotes the unit tangent vector at P
Now, we take an arbitrary point T on the space curve C in which
\(\vec{OT}=R\)
Then, equation of tangent line is
\(\vec{OT}=\vec{OP}+\vec{PT}\)
or
\(R=\vec{r}+\lambda \vec{t}\)
or
\(R=\vec{r}+\lambda \vec{r}'\)
Equation of tangent line in cartesian Form
Let \( C:\vec{r}=\vec{r}( s )\) be a space curve and P be a point on it.
Then, equation of tangent line at P is
\(R=\vec{r}+\lambda \vec{r}'\)
Where
\(\vec{r}\) = (x,y,z)
Diff. \(\vec{r}\) w. r. to. s, we get
\(\vec{r}'=(x',y',z')\)
Also,
\(R=(X,Y,Z)\)
Now, Cartesian form of tangent line is
\(R=\vec{r}+\lambda \vec{r}'\)
or\((X,Y,Z)=(x,y,z)+\lambda (x',y',z')\)
or\((X,Y,Z)=(x+\lambda x',y+\lambda y',z+\lambda z')\)
or\(X=x+\lambda x',Y=y+\lambda y',Z=z+\lambda z'\)
or \(\frac{X-x}{x'}=\lambda ,\frac{Y-y}{y'}=\lambda ,\frac{Z-z}{z'}=\lambda \)
or \(\frac{X-x}{x'}=\frac{Y-y}{y'}=\frac{Z-z}{z'}=\lambda \)
This completes the solution.
The alternative form is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}=\lambda \)
This completes the solution.
Examples
- Find equation of tangent line on \(\vec{r}=( 1+t,-t^2,1+t^3 )\) at t=1.
Solution
Given space curve is
\(\vec{r}=( 1+t,-t^2,1+t^3 )\) (1)
At t=1, the position of \( \vec{r}\) is
\(\vec{r}\)= (x,y,z) = (2,-1,2)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=( 1,-2t,3t^2 )\) (2)
At t=1, the position of \(\dot{\vec{r}}\) is
\(\dot{\vec{r}}=(\dot{x},\dot{y},\dot{z})=( 1,-2,3 )\)
Now, equation of tangent line at t=1 is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-2}{1}=\frac{Y+1}{-2}=\frac{Z-2}{3}\) From (i), and (ii)
This completes the solution. - Find equation of tangent line to circular helix \( \vec{r}=( acost,asint,bt )\).
Solution
Given equation of circular helix is
\(\vec{r}=( acost,asint,bt )\) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=( -asint,acost,b )\) (2)
Now, equation of tangent line on the helix is
\(\frac{X-x}{{\dot{x}}}=\frac{Y-y}{{\dot{y}}}=\frac{Z-z}{{\dot{z}}}=\mu \)
or\(\frac{X-acost}{-asint}=\frac{Y-asint}{acost}=\frac{Z-bt}{b}\)
This completes the solution. - Show that tangent on \(\vec{r}=( 3t,3t^2,2t^3 )\) makes a constant angle with a line \(y=z-x=0\).
Solution
Given equation of space curve is
\(\vec{r}=( 3t,3t^2,2t^3 )\) (1)
Diff. (i) w. r. to. t, we get
\(\vec{r}=( 3,6t,6t^2 )\)
Here, direction ratios of the tangent line is
\(3,6t\) and \(6t^2\)
or \(1,2t\) and \(2t^2\)
Also, given straight line is
\(y=z-x=0\)
The direction ratios of the line is
\(1,0,1\)
If \( \theta \) be the angle between curve and line, then angle between them is
\(\cos \theta =\frac{1.1+2t.0+2t^2.1}{\sqrt{1+0+1}.\sqrt{1+4t^2+4t^4}}=\frac{1}{\sqrt{2}}\)
or\(\theta =\frac{\pi }{4}\) ; which is constant
Thus, curve \( \vec{r}=( 3t,3t^2,2t^3 )\) makes a constant angle with line \( y=z-x=0\).
ExerciseFind the equation of tangent line to the following curev
- \( \vec{r}(t) = (t^2, \cos t, 2t + 5) \) at \( t = 0 \)
Solution
Given equation of space curve is
\(\vec{r}=(t^2, \cos t, 2t + 5) \) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=(2t, - \sin t, 2) \)
The equation of tangent line is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-t^2}{2t}=\frac{Y- \cos t}{- \sin t}=\frac{Z-2t-5}{2}\)
Now, equation of tangent line at t=0 is
\(\frac{X}{0}=\frac{Y- 1}{0}=\frac{Z-5}{2}\)
This completes the solution. - \( \vec{r}(t) = (\sin t, t^2, \cos t + 3) \) at \( t = \pi \)
Solution
Given equation of space curve is
\(\vec{r}=(\sin t, t^2, \cos t + 3) \) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=(\cos t, 2t, - \sin t) \)
The equation of tangent line is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-\sin t}{\cos t}=\frac{Y- t^2}{2t}=\frac{Z-\cos t - 3}{- \sin t}\)
Now, equation of tangent line at t= \(\pi\) is
\(\frac{X}{-1}=\frac{Y- \pi^2}{2\pi}=\frac{Z-2}{0} \)
This completes the solution. - \( \vec{r}(t) = (e^t, t - 1, \log(t + 1)) \) at \( t = 1 \)
Solution
Given equation of space curve is
\(\vec{r}=(e^t, t - 1, \log(t + 1)) \) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=(e^t, 1, \frac{1}{t + 1}) \)
The equation of tangent line is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-e^t}{e^t}=\frac{Y- (t - 1)}{1}=\frac{Z-\log(t + 1)}{\frac{1}{t + 1}}\)
Now, equation of tangent line at t= 1 is
\(\frac{X-e}{e}=\frac{Y}{1}=\frac{Z- \log(2)}{\frac{1}{2}} \)
This completes the solution. - \( \vec{r}(t) = (\cosh t, \sinh t, t^2 - 2) \) at \( t = 0 \)
Solution
Given equation of space curve is
\(\vec{r}=(\cosh t, \sinh t, t^2 - 2) \) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=(\sinh t, \cosh t, 2t) \)
The equation of tangent line is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-\cosh t}{\sinh t}=\frac{Y- \sinh t}{\cosh t}=\frac{Z-t^2 + 2}{2t}\)
Now, equation of tangent line at t= 0 is
\(\frac{X-1}{0}=\frac{Y- 0}{1}=\frac{Z- -2}{0} \)
This completes the solution. - \( \vec{r}(t) = (t^3, \frac{1}{t}, \sin(2t)) \) at \( t = 1 \)
Solution
Given equation of space curve is
\(\vec{r}=(t^3, \frac{1}{t}, \sin(2t)) \) (1)
Diff. (i) w. r. to. t, we get
\(\dot{\vec{r}}=(3t^2, -\frac{1}{t^2}, 2\cos(2t)) \)
The equation of tangent line is
\(\frac{X-x}{\dot{x}}=\frac{Y-y}{\dot{y}}=\frac{Z-z}{\dot{z}}\)
or \(\frac{X-t^3}{3t^2}=\frac{Y-\frac{1}{t}}{-\frac{1}{t^2}}=\frac{Z-\sin(2t)}{2\cos(2t)}\)
Now, equation of tangent line at t= 1 is
\(\frac{X-1}{3}=\frac{Y- 1}{-1}=\frac{Z- \sin(2)}{2\cos(2)} \)
This completes the solution.
Thank you so much Guru 🙏
ReplyDelete