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)b =b3/c3

2. (𝛿𝑝/π›Ώπ‘Ž )A=b/c



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

1. From Pythagoras theorem


c2=a2+b2c^2=a^2+b^2\\


The area of the triangle


A=12ab=12pcA=\dfrac{1}{2}ab=\dfrac{1}{2}pc

Then


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

p=p(a,b)=aba2+b2p=p(a, b)=\dfrac{ab}{\sqrt{a^2+b^2}}

Differentiate by aa


βˆ‚pβˆ‚a=ba2+b2βˆ’a2ba2+b2(a2+b2)\dfrac{\partial p}{\partial a}=\dfrac{b}{\sqrt{a^2+b^2}}-\dfrac{a^2b}{\sqrt{a^2+b^2}(a^2+b^2)}

=a2b+b3βˆ’a2ba2+b2(a2+b2)=\dfrac{a^2b+b^3-a^2b}{\sqrt{a^2+b^2}(a^2+b^2)}

Substitute c=a2+b2c=\sqrt{a^2+b^2}


βˆ‚pβˆ‚a=b3c3\dfrac{\partial p}{\partial a}=\dfrac{b^3}{c^3}

2.


A=const=>12ab=constA=const=>\dfrac{1}{2}ab=const

Let a=kb,k>0.a=kb, k>0. Then c2=a2+b2=k2b2+b2=b2(1+k2).c^2=a^2+b^2=k^2b^2+b^2=b^2(1+k^2).


bc=1k2=const\dfrac{b}{c}=\sqrt{\dfrac{1}{k^2}}=const

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

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


(βˆ‚pβˆ‚a)A=bc(\dfrac{\partial p}{\partial a})_A=\dfrac{b}{c}


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