# The chain rule

## When are two lines perpendicular?

• Suppose given two lines $$L_1:(y-y_1)=m_1(x-x_1)$$, $$L_2:(y-y_2)=m_2(x-x_2)$$.
• As we have seen, the slopes give the average rate of change: $$m_1=\frac{\Delta y_1}{\Delta x_1}$$.
• Therefore, if a right triangle has legs of (signed) length $$\Delta y_1$$ and $$\Delta x_1$$, and the angle agains the side of length $$\Delta y_1$$ is $$\theta_1$$, then we have $$m_1=\tan\theta_1$$.
• Similarly, we get $$m_2=\tan\theta_2$$.
• Note that $$\cot\theta_1=\frac{1}{\tan\theta_1}=\frac{1}{m_1}$$ and $$\cot\theta_2=\frac{1}{m_2}$$.
• The two lines are perpendicular, if the difference of the two angles is $$\frac{\pi}{2}$$. That is, we need $\theta_1-\theta_2=\pm\frac{\pi}{2}.$
• Let's take cotangents on both sides. $\frac{\cot\theta_1\cot\theta_2+1}{\cot\theta_2-\cot\theta_1}=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2}=\cot(\theta_1-\theta_2)=\cot(\pm\frac{\pi}{2})=0.$
• That is, we need $m_1^{-1}m_2^{-1}=\cot\theta_1\cot\theta_2=-1,$ that is $$m_2=-m_1^{-1}$$.

## 3.4.60

• At what point on the curve $$C:y=\sqrt{1+2x}$$ is the tangent line perpendicular to the line $$L:6x+2y=1$$?
• Last time we saw that we can rearrange the equation of the line $$L$$ to $$y-\frac12=-3x$$, so in particular the slope of $$L$$ is $$m_L=-3$$.
• We also calculated the slope of the tangent line at $$x=x_0$$ to $$C$$: $$f'(x=x_0)=(1+2x_0)^{-1/2}$$.
• Now we can answer the original question: for the tangent line at $$x=x_0$$ of $$C$$ to be perpendicular to $$L$$, we need $$3=\sqrt{1+2x_0}$$, that is $$x_0=4$$.
• Exercises. 3.4: 86, 96, 99