Find the points on the cone z^2=x^2+y^2 that are closest to the point (4,2

floymdiT 2021-10-18 Answered
Find the points on the cone z2=x2+y2 that are closest to the point (4,2,0)
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

tabuordy
Answered 2021-10-19 Author has 91 answers

Using the distance formula we can say that the distance between a general point (x,y,z) and (4,2,0) is given by
D=(x4)2+(y2)2+(z0)2
Since the points are on the cone we can replace z2 with x2+y2
D=(x4)2+(y2)2+x2+y2
Consider the following function
f(x,y)=(x4)2+(y2)2+x2+y2
Note that D(x,z) will be minimum when f(x,z) is minimum, so we will work with f(x,z) for conveince
To find the minimum value of f(x,y), we will us the second derivative test.
Second derivatives test:
If the second partial derivatives of f are continuous on a disk containing the critical point (a,b) , and
D(a,b)=fxx(a,b)fyy(a,b)[fxy(a,b)]2
Then we can conclude that:
1) D>0 and fxx(a,b)<0 a local minimum
2) D>0 and fxx(a,b)<0 a local maximum
3) D<0 no local maximuum/minimum
We can find the local extrema using the Second Derivatives Tes from the section. Note that
f(x,y)=(x4)2+(y2)2+x2+y2
is a polynomial function and hence is continuous everywhere in its domain. Thus the theorem applies.
Firstly, we will need the first and second partial derivatives of f:
fx(x,y)=2(x4)+2x fy(x,y)=2(y2)+2y
 fxx(x,y)=2(1)+2(1)=4 fyx(x,y)=0
fxy(x,y)=0 fyy(x,y)=2(1)+2(1)=4
Now we find the critical points of f. These are the points where both first partial derivatives are 0 at the same time. So let's solve the system of equations
{2(x4)+2x=02(y2)+2y=0
Solve the first equation
(x4)+x=0
2x=4
Divide both sides by 2
x=2
Solve the first equation
2(y2)+2y=0
y2+y=0
2y=2
Divide both sides by 2
y=1
Therefore, the critical point is (2,1)
Now we just neet the value

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more