using the chain rule find the min and max points and their values of the composite function defined by z=x^2+y^2, x=sin(2t), y=cos(t)
This is what Ive got so far...
dz/dt= 2x*dx/dt+2y*dy/dt
=4sin2tcos2t-2sintcost
I understand the steps of computing the max and min I just can't seem to solve this problem. Its on a test I have due tomorrow so please help!!
You need to find dz/dt=0, so set
4sin2tcos2t-2sintcost =0
Expand using double-angle formulae:
8cos(t)sin(t)(cos(t)^2-sin(t)^2)-2cos(t)sin(t) = 0
Factorize:
-2cos(t)sin(t)(4sin(t)^2-4cos(t)^2+1)
Which gives
t=0, t=π/2, t=π and t=3π/2
when sin(t)=0 or cos(t)=0.
This leaves us with:
4sin(t)^2-4cos(t)^2+1=0
which can be reduced by the substitution
cos(t)^2=1-sin(t)^2
to a quadratic equation equivalent to
sin(t)=sqrt(3/8)
Now that leaves you to check each point as a maximum/minimum or inflection point using second derivatives.
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To find the critical points of the composite function defined by z=x^2+y^2, with x=sin(2t) and y=cos(t), we need to find the values of t that make dz/dt equal to zero.
You have already correctly found dz/dt to be equal to 4sin(2t)cos(2t) - 2sin(t)cos(t). To find the zeros of dz/dt, we can set it equal to zero and solve the equation:
4sin(2t)cos(2t) - 2sin(t)cos(t) = 0
Now, we can factor out a common term of 2sin(t)cos(t) from both terms:
2sin(t)cos(t)(2cos(2t) - 1) = 0
Since 2sin(t)cos(t) is never zero, the equation reduces to:
2cos(2t) - 1 = 0
Now, solve for t:
2cos(2t) = 1
cos(2t) = 1/2
Using the inverse cosine function, we find that:
2t = ±π/3 + 2πn or 2t = ±5π/3 + 2πn
where n is an integer. Divide both sides of the equation by 2 to solve for t:
t = ±π/6 + πn or t = ±5π/6 + πn
These are the values of t that correspond to the critical points of the composite function.
To find the corresponding (x, y) values and the values of the composite function at these critical points, substitute the values of t into the equations for x and y:
For t = π/6, we have:
x = sin(2(π/6)) = sin(π/3) = √3/2
y = cos(π/6) = √3/2
For t = -π/6, we have:
x = sin(2(-π/6)) = sin(-π/3) = -√3/2
y = cos(-π/6) = √3/2
For t = 5π/6, we have:
x = sin(2(5π/6)) = sin(5π/3) = -√3/2
y = cos(5π/6) = -√3/2
For t = -5π/6, we have:
x = sin(2(-5π/6)) = sin(-5π/3) = √3/2
y = cos(-5π/6) = -√3/2
Now, substitute these (x, y) values into the function z = x^2 + y^2 to find the values of the composite function at these critical points:
For t = π/6, we have:
z = (√3/2)^2 + (√3/2)^2 = 3/4 + 3/4 = 3/2
For t = -π/6, we have:
z = (-√3/2)^2 + (√3/2)^2 = 3/4 + 3/4 = 3/2
For t = 5π/6, we have:
z = (-√3/2)^2 + (-√3/2)^2 = 3/4 + 3/4 = 3/2
For t = -5π/6, we have:
z = (√3/2)^2 + (-√3/2)^2 = 3/4 + 3/4 = 3/2
Therefore, the min and max points of the composite function are:
(√3/2, √3/2), (-√3/2, √3/2), (-√3/2, -√3/2), and (√3/2, -√3/2)
And the corresponding values of the composite function at these points are:
3/2, 3/2, 3/2, and 3/2, respectively.
Please note that verifying these critical points as minima or maxima requires additional tests, such as the second derivative test or using inequalities.