We can use derivative to find the tangents and normals of continuous curves [Continuity]. We know that the slope of a tangent to a continuous curve $y=ƒ(x)$ is given by \[\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0} \frac{ƒ(x+\Delta x)-ƒ(x)}{\Delta x}\]\[=\frac{dy}{dx}\]
The detailed explanation of drawing a tangent line to the curve is given in this page: Tangent Line to a Curve (Origin of Differential Calculus). Here, we will discuss about finding the tangents and normals of different functions.
Equations of Tangent and Normal
Let a function $y=ƒ(x)$ have a finite derivative $ƒ'(x_1)$ at $x=x_1$. Then, the curve $y=ƒ(x)$ has a tangent at a point $P(x_1,y_1)$ whose slope or gradient is, \[m=\tan\theta=ƒ'(x)=\left(\frac{dy}{dx}\right)_{x=x_1}\]
The equation of the tangent is \[y-y_1=m(x-x_1)\]
If $m=0$, the curve has a tangent parallel to the $\text{X-axis}$ at that point; its equation is $y=y_1$ and it is known as horizontal tangent. If $y=ƒ(x)$ is continuous at $x=x_1$ but $\lim_{x\to x_1} ƒ'(x)=\infty$, the curve has a tangent parallel to $\text{Y-axis}$ at that point; its equation is $x=x_1$ and it known as vertical tangent.
The normal to a curve at any point $P(x_1,y_1)$ is a line through the point which is perpendicular to the tangent at that point.
The equation to the normal at $P(x_1,y_1)$ is \[y-y_1=-\frac{1}{m}(x-x_1)\] \[\begin{array}{c} \text{If the normal is horizontal, }& x=x_1 \\ \text{If the normal is vertical, }& y=y_1\end{array}\]
The angle of intersection of two curves
If two curves intersect, the angle between the tangents to the curve at their point of intersection is the angle between the curves at the point.
The angle of intersection $\phi$ between the two curves is given by, \[\tan\phi=\frac{m_1-m_2}{1+m_1m_2}\]
If $m_1=m_2$, the angle of intersection is $\phi=0°$ and if $m_1=-\frac{1}{m_2}$, the angle of intersection is $\phi=90°$.
Find the slope and the inclination with the $\text{X-axis}$ of the tangent of $x^2+y^2=25$ at $(-3, 4)$.
Here, \[x^2+y^2=25\] Differentiating with respect to $x$, \[2x+2y\frac{dy}{dx}=0\] \[\frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}\] At $(-3,4)$, \[\frac{dy}{dx}=\frac{3}{4}\] \[\therefore \text{Slope (m)}=\frac{3}{4}\] \[\therefore \tan\theta=\frac{3}{4}\] \[\theta=\tan^{-1}\left(\frac{3}{4}\right)\]
At what angle does the curve $y(1+x)=x$ cut the $\text{X-axis}$?
Given curve is, \[y(1+x)=x\] The curve cuts the $\text{X-axis}$ where $y=0$, then, $x=0$. Now, \[y=\frac{x}{1+x}\] Differentiating with respect to $x$, \[\frac{dy}{dx}=\frac{(1+x)\frac{dx}{dx}-x\frac{d(1+x)}{dx}}{(1+x)^2}\] \[=\frac{1+x-x(0+1)}{(1+x)^2}\] \[=\frac{1}{(1+x)^2}\]
At $x=0$, \[\frac{dy}{dx}=\frac{1}{(1+0)^2}=1\] \[\therefore \tan\theta=1\] \[\tan\theta=\tan\frac{π}{4}\] \[\therefore \theta=\frac{π}{4}\]
Find the equations of the tangents and normals to the curve $y=2x^3-5x^2+8$ at $(2,4)$.
Given curve is, \[y=2x^3-5x^2+8\] Differentiating with respect to $x$, \[\frac{dy}{dx}=6x^2-10x\] At $(2,4)$, \[\frac{dy}{dx}=6×2^2-10×2=4\] \[\therefore \text{Slope }(m)=4\]
The equation of tangent to the curve at $(2,4)$ is, \[y-y_1=m(x-x_1)\] \[y-4=4(x-2)\] \[y-4=4x-8\] \[4x-y-4=0\]
Again, equation of normal at $(2,4)$ is, \[y-y_1=\frac{-1}{m}(x-x_1)\] \[y-4=\frac{-1}{4}(x-2)\] \[4y-16=-x+2\] \[\therefore x+4y-18=0\]
Find the points on the curve $y=x^3-3x^2+1$ where the tangents are parallel to the $\text{X-axis}$.
Given curve is, \[y=x^3-3x^2+1\] Differentiating with respect to $x$, \[\frac{dy}{dx}=3x^2-6x\]
For the tangent parallel to $\text{X-axis}$, \[\frac{dy}{dx}=0\] \[3x^2-6x=0\] \[3x(x-2)=0\] \[\text{Either, } x=0\text{ or }x=2\] \[\text{If }x=0, \text{ then }y=1.\] \[\text{If }x=2, \text{ then } y=2^3-3×2^2+1=-3.\]
Hence, the required points are $(0,1)$ and $(2,-3)$.
Find the points on the circle $x^2+y^2=16$ at which the tangents are parallel to the $\text{Y-axis}$.
Given circle is, \[x^2+y^2=16\] Differentiating with respect to $x$, \[2x+2y\frac{dy}{dx}=0\] \[2y\frac{dy}{dx}=-2x\] \[\therefore \frac{dy}{dx}=\frac{-x}{y}\]
For the tangent parallel to $\text{Y-axis}$, \[\frac{dx}{dy}=0\] \[\frac{-y}{x}=0\] \[y=0\] At $y=0$, $x^2=16\Rightarrow x=\pm 4$. Hence, the required points are $(4,0)$ and $(-4,0)$.
Find the point on the curve $y^2=4x+1$ at which the tangent is perpendicular to the line $7x+2y=10$.
Given curve is, \[y^2=4x+1\] Differentiating with respect to $x$, \[2y\frac{dy}{dx}=4\] \[\therefore \frac{dy}{dx}=\frac{2}{y}\] \[\therefore \text{Slope of the curve }(m_1)=\frac{2}{y}\]
Also, given line is, \[7x+2y=10\] Differentiating with respect to $x$, \[7+2\frac{dy}{dx}=0\] \[\frac{dy}{dx}=\frac{-7}{2}\] \[\therefore \text{Slope of the line }(m_2)=\frac{-7}{2}\]
Since the tangent is perpendicular to the line, we have, \[m_1m_2=-1\] \[\frac{2}{y}.\frac{-7}{2}=-1\] \[\therefore y=7\] At $y=7$, \[7^2=4x+1\] \[\therefore x=12\] Hence, the required point is $(12,7)$.
Show that the equation of the tangent to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(a,b)$ is $\frac{x}{a}+\frac{y}{b}=2$.
Given curve is, \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] Differentiating with respect to $x$, \[\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\] \[\frac{2y}{b^2}\frac{dy}{dx}=-\frac{2x}{a^2}\] \[\therefore \frac{dy}{dx}=\frac{-b^2x}{a^2y}\] \[\therefore \text{Slope }(m)=\frac{-b^2x}{a^2y}\] At $(a,b)$, \[m=\frac{-b^2a}{a^2b}=\frac{-b}{a}\]
Equation of tangent to the curve at the point $(a,b)$ is, \[y-y_1=m(x-x_1)\] \[y-b=\frac{-b}{a}(x-a)\] \[a(y-b)=-b(x-a)\] \[ay-ab=-bx+ab\] \[ay+bx=2ab\] \[\therefore \frac{x}{a}+\frac{y}{b}=2\]
Find the angle of intersection of the curves $y=6-x^2$ and $x^2=2y$.
Here, \[y=6-x^2\text{ __(1)}\] Differentiating $\text{(1)}$ with respect to $x$, \[\frac{dy}{dx}=-2x\] also, \[x^2=2y\text{ __(2)}\] Differentiating $\text{(2)}$ with respect to $x$, \[2x=2\frac{dx}{dy}\] \[\frac{dy}{dx}=x\] Solving $\text{(1)}$ and $\text{(2)}$, \[x^2=2(6-x^2)\] \[x^2=12-2x^2\] \[3x^2=12\] \[x^2=4\] \[x=\pm 2\] \[\text{When }x=2, y=6-2^2=2\] \[\text{When }x=-2, y=6-(-2)^2=2\] Hence, the points of intersection are $(2,2)$ and $(-2,2)$. At $(2,2)$, \[\text{Slope of }(y=6-x^2)\text{ }(m_1)=-2×2=-4\] \[\text{Slope of }(x^2=2y)\text{ }(m_2)=2\] Now, \[\tan\theta=\frac{m_1-m_2}{1+m_1m_2}\] \[\tan\theta=\frac{-4-2}{1+(-4)2}=\frac{6}{7}\] \[\therefore \theta=\tan^{-1}\left(\frac{6}{7}\right)\] At $(-2,2)$, \[\text{Slope of }(y=6-x^2)\text{ }(m_3)=-2×-2=4\] \[\text{Slope of }(x^2=2y)\text{ }(m_4)=-2\] Now, \[\tan\theta=\frac{m_3-m_4}{1+m_3m_4}=\frac{4+2}{1+4×(-2)}=-\frac{6}{7}\] \[\therefore \theta=\tan^{-1}\left(-\frac{6}{7}\right)\]