Answer to Question #227749 in Calculus for Anuj

Question #227749

Question:

If π‘Ž 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/π›Ώπ‘Ž)bΒ =b3/c3

2. (𝛿𝑝/π›Ώπ‘Ž )A=b/c



1
Expert's answer
2021-08-20T11:43:55-0400

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}"


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