- #1
Tori No Otoko
- 1
- 0
Slightly confused at what it wants me to do here?
MHB Derivatives with Quadratic Functions.
- Thread starter Tori No Otoko
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The discussion focuses on solving a derivative equation involving quadratic functions. The key equation presented is T'(z) = 0, leading to a series of algebraic manipulations to simplify the expression. The final form reveals a quadratic equation in z, specifically 39z^2 - 128zD + 64D^2 = 0. It is emphasized that differentiation is not needed; instead, the problem requires setting the derivative equal to zero and performing algebraic operations. The main takeaway is the importance of recognizing the problem's requirements and applying algebraic techniques to reach the solution.
Slightly confused at what it wants me to do here?
- #2
skeeter
- 1,103
- 1
just an algebra drill ...
$T'(z) = 0 \implies \dfrac{z}{\sqrt{80^2+z^2}} = \dfrac{D-z}{\sqrt{50^2+(z-D)^2}}$
$z\sqrt{50^2+(z-D)^2} = (D-z)\sqrt{80^2+z^2}$
$z^2[50^2+(z-D)^2] = (D-z)^2[80^2+z^2]$
$50^2z^2 + z^2(z-D)^2 = 80^2(D-z)^2 + z^2(D-z)^2$
$0 = 80^2(D-z)^2 - 50^2z^2$
$0 = 80^2(D^2 - 2zD + z^2) - 50^2z^2$
$0 = 80^2D^2 - 80^2 \cdot 2zD + (80^2-50^2)z^2$
$0 = 6400(D^2-2zD) + 3900z^2$
$0 = 64(D^2-2zD) + 39z^2$
$0 = 39z^2 - 128zD + 64D^2$
$T'(z) = 0 \implies \dfrac{z}{\sqrt{80^2+z^2}} = \dfrac{D-z}{\sqrt{50^2+(z-D)^2}}$
$z\sqrt{50^2+(z-D)^2} = (D-z)\sqrt{80^2+z^2}$
$z^2[50^2+(z-D)^2] = (D-z)^2[80^2+z^2]$
$50^2z^2 + z^2(z-D)^2 = 80^2(D-z)^2 + z^2(D-z)^2$
$0 = 80^2(D-z)^2 - 50^2z^2$
$0 = 80^2(D^2 - 2zD + z^2) - 50^2z^2$
$0 = 80^2D^2 - 80^2 \cdot 2zD + (80^2-50^2)z^2$
$0 = 6400(D^2-2zD) + 3900z^2$
$0 = 64(D^2-2zD) + 39z^2$
$0 = 39z^2 - 128zD + 64D^2$
- #3
HOI
- 921
- 2
Notice that the problem gives you T', not T. It is not necessary to differentiate. Just set it equal to 0 and do the algebra to get the final result.
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