Got a questions that i not sure how to do:
Could someone help me with that.Code:Find the equation of the tangent plane to the surface z+32 = (x^5)(e^4) ycos( z ) at the point (2, 0, 0).
Thanks
$\displaystyle z+32=(x^5)(e^4)y\cos(z) \iff 0=(x^5)(e^4)y\cos(z)-z-32$
$\displaystyle f(x,y,z)=(x^5)(e^4)y\cos(z)-z-32$ to find the normal vector we take the gradient
$\displaystyle \nabla f = 5x^4e^4y\cos(z) \vec i +x^5e^4\cos(z) \vec j -(x^5e^4y\sin(z)-1) \vec k$
$\displaystyle \nabla f (2,0,0)=0 \vec i +32e^4 \vec j - 1\vec k$
let (x,y,z) be in the plane then a vector in the plane (x-2,y,z)
doting these two vectors and setting equal to zero will give the equation of the plane.
$\displaystyle 0(x-2)+32e^4(y)+(-1)(z)=0 \iff z=32e^4y$
Are you sure its right cos i tested it at this online solution thing and it said it wasnt:
It said:
Hint: the equation of a plane, in general, is given by expanding
r •n = r0 •n
where r = [x, y, z], r0 is the position vector of a point on the plane, and n is a normal to the plane.
Now, rearrange the given equation in the form f(x, y, z) = 0. Then ∇f has the property of being normal to the surface at any point on the surface defined by f(x, y, z) = 0. Thus ∇f evaluated at (2, 0, 0) and (2, 0, 0) itself, are respectively, a normal and a point on the tangent plane you seek.
If you mean this equation...
$\displaystyle z+32=(x^5)(e^{4y})\cos(z) \iff 0=(x^5)(e^{4y})\cos(z)-z-32$
$\displaystyle f(x,y,z)=(x^5)(e^{4y})\cos(z)-z-32$ to find the normal vector we take the gradient
$\displaystyle \nabla f = 5x^4e^{4y}\cos(z) \vec i +4x^5e^{4y}\cos(z) \vec j -(x^5e^{4y}\sin(z)+1) \vec k$
$\displaystyle \nabla f (2,0,0)=80 \vec i +128 \vec j - 1\vec k$
let (x,y,z) be in the plane then a vector in the plane (x-2,y,z)
doting these two vectors and setting equal to zero will give the equation of the plane.
$\displaystyle 80(x-2)+128(y)+(-1)(z)=0 \iff 80x-160+128y-z=0 \iff$
$\displaystyle 80x+128y-z=160$