Find the value of a
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I am trying to solve this in MATLAB.
Find the value of the number a such that the families of curves y = (x + c)-1 and y= a (x + k)1/3 are orthogonal trajectories.
Here are the steps I took to solve the problem however I can't figure out how to do this in MATLAB.
Step 1.
y’= -1(x + c)-2(1) = -1(1 / (x + c)2
Step 2.
y’= - 1/(x0+c)2
Step 3.
y’= a/3 (x + k)-2/3 (1) = a / 3(x + k)2/3
Step 4.
y’= a/ 3(x0+k)2/3
Step 5
y = a(-k+k)1/3 = a(0) = 0
Step 6
(- 1/(x0+c)2) (a/3(x0+k)2/3) = -1, a = 3(x0+c)2(x0+k)2/3
Step 7
(x0+c)-1= a(x0+k)1/3, 1/(x0+c) = a(x0+k)1/3, (x0+c) = 1/a(x0+k)1/3
Step 8
a = 3(1/a(x0+k)1/3)2(x0+k)2/3
a = 3(1/a2(x0+k)2/3) (x0+k)2/3
a = 3(1/a2)
a3 = 3 a = 3sqrt(3)
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Réponses (2)
Carlos Guerrero García
le 19 Nov 2022
Hi Trenton...!!!
You can find my suggestion for an animation plot in
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