invert the function s = L(t) to solve for t.

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Hyunji Yu 2020-3-9

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回答： SAI SRUJAN 2024-5-30

syms t;
x(t)= sin(3*t^2)*(12*t + (10*13^(1/2))/13);
y(t)= t*(6*13^(1/2)*t + 5);
z(t)= cos(3*t^2)*(12*t + (10*13^(1/2))/13);
syms tau;
L(t) = vpaintegral(speed(tau), tau, 0, t);
syms s;
solve(s == L(t), t);

I'm trying to invert the function s = L(t) to solve for t, but I don't know how to change the function regarding as t.

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Nishant Gupta 2020-3-13

Can you please tell what is speed(tau)?

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Answer 1

SAI SRUJAN 2024-5-30

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Hi Hyunji,

I understand that you are trying to invert the function 's=L(t)' to solve for 't'.

The speed '(v(t))' of a particle moving along a path in three-dimensional space is given by the magnitude of its velocity vector, which is the derivative of its position vector, then: ['v(t) = sqrt(dx^2 + dy^2 + dz^2);'].

Please go through the following code sample to proceed further,

syms s t tau;
x(t) = sin(3*t^2)*(12*t + (10*sqrt(13))/13);
y(t) = t*(6*sqrt(13)*t + 5);
z(t) = cos(3*t^2)*(12*t + (10*sqrt(13))/13);
dx = diff(x, t);
dy = diff(y, t);
dz = diff(z, t);
% Speed function v(t)
v(t) = sqrt(dx^2 + dy^2 + dz^2);
% Define L(t) as the integral of v(tau) from 0 to t
L(t) = int(v(tau), tau, 0, t);
tSol = vpasolve(s == L(t), t);