Tangents and derivatives

Consider two curves with equations $$y = (4x+1)^{0.5}$$ and $$y = 0.5x^2 +1$$. The two curves intersect at point $$Q(2,3)$$ and point $$P(0,1)$$. The tangents to each curve at the point $$Q$$ form an angle $$A$$. Find the value of $$A$$.

I saw a formula on the web which says $$\tan A = \frac{|m_1-m_2|}{|1 + m_1m_2|}$$. When using this equation to solve for $$A$$ I get the correct answer. However, I don’t want to use this formula blindly and cannot see a well written proof on the web. I am looking for a derivation of that equation or a different approach to solve that particular problem will do good.

• I think $m_1$ and $m_2$ are the slopes of each of the curves at the intersection point, right? – Matti P. Apr 19 at 11:20
• m1 and m2 are the slopes of the tangent at point Q – Anay Chadha Apr 19 at 11:21
• Hint: It comes from the compound angle formula $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$. – Tony Ip Apr 19 at 11:51

The derivation of that equation is based on the relation $$\tan \theta = m$$ where $$m$$ is the slope (gradient) of a line and $$\theta$$ is the angle the line makes with the horizontal axis. That relation follows immediately from the definition of the slope as "rise" over "run" and comparing that to "opposite" over "adjacent" of an appropriate right triangle.

On that basis we can state that $$\theta = \arctan m$$.

The angle between two lines is the absolute value of the difference between the angles each line makes to a common reference line (for convenience, take the horizontal axis). So the required angle $$A$$ is $$|\arctan m_1 - \arctan m_2|$$

So $$\tan A = \tan |\arctan m_1 - \arctan m_2|$$

Now apply the tangent of sum/difference identity (i.e. $$\tan (A \pm B) = \frac{\tan A \pm \tan B} {1 \mp \tan A \tan B}$$) to simplify that. You may find it convenient to split up the cases $$\arctan m_1 \geq \arctan m_2$$ and $$\arctan m_1 < \arctan m_2$$ but you can express the final result simply with the absolute value notation.

See $$m_1$$ and $$m_2$$ are nothing but slopes of the two lines and the angle between them, and you may know that the slope of a line is simply defined by $$\frac{Δy}{Δx}$$. Now if the line makes an angle $$\theta$$ with the x-axis then, $$m=\frac{Δy}{Δx}=tan{\theta}$$. You can see the image here for reference for the above relation

Now since the angle formed between the two line is nothing but the difference between the angles they make with the x-axis ie if $$L_1$$ and $$L_2$$ be two lines making angles $$\theta_1$$ and $$\theta_2$$ with the x axis respectively then the angle between those two lines would be nothing but $${\theta_1}-{\theta_2}$$. Now for the formula,$${\theta_1}-{\theta_2}=\arctan{m_1}-\arctan{m_2}$$

$$\implies {\theta_1}-{\theta_2}=\arctan\left({\frac{{m_1}-{m_2}}{1+{m_1}{m_2}}}\right)$$

If we assume that difference is $$\theta$$, $$\theta=\arctan\left({\frac{{m_1}-{m_2}}{1+{m_1}{m_2}}}\right)$$ $$\implies \tan{\theta}=\frac{{m_1}-{m_2}}{1+{m_1}{m_2}}$$

That modulus is there because we may not know which of the slopes is greater so while taking the difference of angles we just take the absolute value because of the argument of the $$\arctan{\theta}$$.