Re: Ivan's question at Yahoo! Answers regarding finding the line tangent to an implicitly defined ci
Hello Ivan,
We are given the implicitly defined curve:
$$\ln(xy)+2x-y+1=0$$
and are asked to find the equation of the line tangent to this curve at the point:
$$(x,y)=\left(\frac{1}{2},2 \right)$$
The first thing I like to do is verify that in fact the given point lies on the curve. So, if we substitute for $x$ and $y$ into the equation of the curve, we find:
$$\ln\left(\frac{1}{2}\cdot2 \right)+2\left(\frac{1}{2} \right)-2+1=0$$
$$\ln(1)+1-2+1=0$$
$$0+1-2+1=0$$
$$0=0$$
So, the given point is on the curve. Next, let's use the logarithmic property:
$$\log_a(bc)=\log_a(b)+\log_a(c)$$
to rewrite the equation of the curve to make implicit differentiation easier:
$$\ln(x)+\ln(y)+2x-y+1=0$$
Now, implicitly differentiate with respect to $x$:
$$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}+2-\frac{dy}{dx}=0$$
We want to solve for $$\frac{dy}{dx}$$, so let's move all terms not involving this to the right side:
$$\frac{1}{y}\frac{dy}{dx}-\frac{dy}{dx}=-\frac{1}{x}-2$$
Multiply through by $-1$ and factor out $$\frac{dy}{dx}$$ on the left side:
$$\frac{dy}{dx}\left(1-\frac{1}{y} \right)=\frac{1}{x}+2$$
Combine terms:
$$\frac{dy}{dx}\left(\frac{y-1}{y} \right)=\frac{2x+1}{x}$$
Multiply through by $$\frac{y}{y-1}$$:
$$\frac{dy}{dx}=\frac{(2x+1)y}{x(y-1)}$$
To get the slope of the tangent line, we need to evaluate this for:
$$(x,y)=\left(\frac{1}{2},2 \right)$$
Hence:
$$m=\left.\frac{dy}{dx}\right|_{(x,y)=\left(\frac{1}{2},2 \right)}=\frac{\left(2\left(\frac{1}{2} \right)+1 \right)2}{\frac{1}{2}(2-1)}=\frac{4}{\frac{1}{2}}=8$$
Now, we have the slope, and a point on the line, so using the point-slope formula, we find the tangent line is given by:
$$y-2=8\left(x-\frac{1}{2} \right)$$
Arranging this in slope-intercept form, we obtain:
$$y=8x-2$$
