Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=t√,y=t2−2t;t=4
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Kevin Thongkham
on 26 Aug 2021
Answered: Kevin Thongkham
on 27 Aug 2021
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=√t, y=t^2−2t; t=4
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Wan Ji
on 26 Aug 2021
Edited: Wan Ji
on 26 Aug 2021
syms t x0(t) y0(t) x y
x0 = sqrt(t); % parametric equation for x
y0 = t^2-2*t; % parametric equation for y
dx = diff(x0); % dx/dt
dy = diff(y0); % dy/dt
eq = subs(dy,t,4)*(x-subs(x0,4)) - subs(dx,t,4)*(y-subs(y0,4)) % this is the line eqaution eq=0
The answer then becomes
eq =
6*x - y/4 - 10
So 6*x - y/4 - 10 = 0 is the equation of the tangent to the curve at t = 4.
You can also solve it to extract y (in that case, the slope should not be inf)
y = solve(eq, y)
y =
24*x - 40
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