Answer to Question #227749 in Calculus for Anuj

Question

Question:

If 𝑎 and 𝑏 are two sides of a right angled hypotenuse 𝑐 and let 𝑝 be the perpendicular from the opposite

vertex on the hypotenuse, show that:

1. (𝛿 𝑝/𝛿𝑎)_b =b³/c³

2. (𝛿𝑝/𝛿𝑎 )_A=b/c

Answer 1

1. From Pythagoras theorem

$$c^2=a^2+b^2\\$$

The area of the triangle

$$A=\dfrac{1}{2}ab=\dfrac{1}{2}pc$$

Then

$$p=\dfrac{ab}{c}$$

$$p=p(a, b)=\dfrac{ab}{\sqrt{a^2+b^2}}$$

Differentiate by $a$

$$\dfrac{\partial p}{\partial a}=\dfrac{b}{\sqrt{a^2+b^2}}-\dfrac{a^2b}{\sqrt{a^2+b^2}(a^2+b^2)}$$

$$=\dfrac{a^2b+b^3-a^2b}{\sqrt{a^2+b^2}(a^2+b^2)}$$

Substitute $c=\sqrt{a^2+b^2}$

$$\dfrac{\partial p}{\partial a}=\dfrac{b^3}{c^3}$$

2.

$$A=const=>\dfrac{1}{2}ab=const$$

Let $a=kb, k>0.$ Then $c^2=a^2+b^2=k^2b^2+b^2=b^2(1+k^2).$

$$\dfrac{b}{c}=\sqrt{\dfrac{1}{k^2}}=const$$

$$p=\dfrac{b}{c}(a)$$

Use that $\dfrac{b}{c}=const,$ if $A=const.$ Then

$$(\dfrac{\partial p}{\partial a})_A=\dfrac{b}{c}$$