PARTIAL DERIVATIVES
379
The plane through (1,1,1) and parallel to the Jtz-plane is y = l. The slope of the tangent line to the
resulting curve is dzldx = 6x = 6. The plane through (1,1,1) and parallel to the yz-plane is x = 1. The
slope of the tangent line to the resulting curve is dzldy = 8y = 8.
42.29
Find equations of the tangent line at the point (—2,1,5) to the parabola that is the intersection of the surface
z = 2x
2 - 3y
2
and the plane y-l.
The slope of the tangent line is dzldx = 4x= —8. Hence, a vector in the direction of the tangent line is
(1,0, -8). (This follows from the fact that there is no change in y and, for a change of 1 unit in x, there is a
change of dzldx in z.) Therefore, a system of parametric equations for the tangent line is x = 2 + t, y = 1,
z = 5-8t (or, equivalently, we can use the pair of planes y = I and 8x + z = -ll).
42.30
Find equations of the tangent line at the point (—2,1,5) to the hyperbola that is the intersection of the surface
z = 2x
2 — 3y
2
and the plane z = 5.
Think of y as a function of x and z. Then the slope of the tangent line to the curve is dyldx. By implicit
differentiation with respect to x, 0 = 4x - dy(dyldx). Hence, at (-2,1,5), 0 = -8 - 6(dyldx), dyldx = -\.
So, a vector in the direction of the tangent line is (1, - f, 0), and parametric equations for the tangent line are
x=— 2+t, y = l— ff, z = 5 (or, equivalently, the pair of planes 4x + 3y = — 5 and z = 5).
42.31
Show that the tangent lines of Problems 42.29 and 42.30 both lie in the plane Sx + 6y + z + 5 = 0. [This is the
tangent plane to the surface z = 2x
2 - 3y
2
at the point (-2,1,5).]
Both lines contain the point (-2,1,5), which lies in the plane ty: 8x + 6y + z + 5 = 0, since 8(-2) +
6(1)+ 5+ 5 = 0. A normal vector to the plane is A = (8,6,1). [A is a surface normal at (-2,1,5).] The
tangent line of Problem 42.29 is parallel to the vector B = (l,0, —8), which is perpendicular to A [since
A • B = 8(1) + 6(0) + l(-8) = 0]. Therefore, that tangent line lies in the plane &. Likewise, the tangent line of
Problem 42.30 is parallel to the vector C = (l, -j,0), which is perpendicular to A [since A-C = 8(1) +
6(— |) + 1(0) = 0]. Hence, that tangent line also lies in plane 9. (We have assumed here the fact that any line
containing a point of a plane and perpendicular to a normal vector to the plane lies entirely in the plane.)
42.32
The plane y — 3 intersects the surface z = 2x
2 + y
2
in a curve. Find equations of the tangent line to this
curve at the point (2,3,17).
The slope of the tangent line is the derivative dzldx = 4x = 8. Hence, a pair of equations for the tangent
line is (z - 17) l(x - 2) = 8, y = 3, or, equivalently, z = 8x + 1, y = 3.
42.33 The plane x = 3 intersects the surface z = x
2 l( y
2 - 3) in a curve. Find equations of the tangent line to this
curve at (3,2,9).
The slope of the tangent line is
Hence, a pair of equations for the tangent
line is (z-9)/(y-2) =-36 and x = 3, or, equivalently, z--36y + sl, x = 3. [Another method is
to use the vector (0,1,—36), parallel to the line, to form the parametric equations x = 3, y = 2+t,
z = 9 - 36/.1
42.34
State a set of conditions under which the mixed partial derivatives f xy (x 0 , y a ) and f yx (x 0 , y 0 ) are equal.
If (x 0 , y 0 ) is inside an open disk throughout which f xy and/^ exist, and if f xy andf yx are continuous at (jc 0 , y 0 ),
then f xy (x 0 , y 0 ) = f yx (x 0 , y 0 ). Similar conditions ensure equality for n>3 partial differentiations, regardless of the order in which the derivatives are taken.
42.35 For f(x, y) = 3x2y - 2xy + 5y2, verify that fxy=fyic.
f x = 6xy-2y, fxy=6x-2. fy =3x2-2x + Wy, fyx = 6x-2.
42.36 For f(x, y) = x7 In y + sinxy, verify fxy=fyx.
379
The plane through (1,1,1) and parallel to the Jtz-plane is y = l. The slope of the tangent line to the
resulting curve is dzldx = 6x = 6. The plane through (1,1,1) and parallel to the yz-plane is x = 1. The
slope of the tangent line to the resulting curve is dzldy = 8y = 8.
42.29
Find equations of the tangent line at the point (—2,1,5) to the parabola that is the intersection of the surface
z = 2x
2 - 3y
2
and the plane y-l.
The slope of the tangent line is dzldx = 4x= —8. Hence, a vector in the direction of the tangent line is
(1,0, -8). (This follows from the fact that there is no change in y and, for a change of 1 unit in x, there is a
change of dzldx in z.) Therefore, a system of parametric equations for the tangent line is x = 2 + t, y = 1,
z = 5-8t (or, equivalently, we can use the pair of planes y = I and 8x + z = -ll).
42.30
Find equations of the tangent line at the point (—2,1,5) to the hyperbola that is the intersection of the surface
z = 2x
2 — 3y
2
and the plane z = 5.
Think of y as a function of x and z. Then the slope of the tangent line to the curve is dyldx. By implicit
differentiation with respect to x, 0 = 4x - dy(dyldx). Hence, at (-2,1,5), 0 = -8 - 6(dyldx), dyldx = -\.
So, a vector in the direction of the tangent line is (1, - f, 0), and parametric equations for the tangent line are
x=— 2+t, y = l— ff, z = 5 (or, equivalently, the pair of planes 4x + 3y = — 5 and z = 5).
42.31
Show that the tangent lines of Problems 42.29 and 42.30 both lie in the plane Sx + 6y + z + 5 = 0. [This is the
tangent plane to the surface z = 2x
2 - 3y
2
at the point (-2,1,5).]
Both lines contain the point (-2,1,5), which lies in the plane ty: 8x + 6y + z + 5 = 0, since 8(-2) +
6(1)+ 5+ 5 = 0. A normal vector to the plane is A = (8,6,1). [A is a surface normal at (-2,1,5).] The
tangent line of Problem 42.29 is parallel to the vector B = (l,0, —8), which is perpendicular to A [since
A • B = 8(1) + 6(0) + l(-8) = 0]. Therefore, that tangent line lies in the plane &. Likewise, the tangent line of
Problem 42.30 is parallel to the vector C = (l, -j,0), which is perpendicular to A [since A-C = 8(1) +
6(— |) + 1(0) = 0]. Hence, that tangent line also lies in plane 9. (We have assumed here the fact that any line
containing a point of a plane and perpendicular to a normal vector to the plane lies entirely in the plane.)
42.32
The plane y — 3 intersects the surface z = 2x
2 + y
2
in a curve. Find equations of the tangent line to this
curve at the point (2,3,17).
The slope of the tangent line is the derivative dzldx = 4x = 8. Hence, a pair of equations for the tangent
line is (z - 17) l(x - 2) = 8, y = 3, or, equivalently, z = 8x + 1, y = 3.
42.33 The plane x = 3 intersects the surface z = x
2 l( y
2 - 3) in a curve. Find equations of the tangent line to this
curve at (3,2,9).
The slope of the tangent line is
Hence, a pair of equations for the tangent
line is (z-9)/(y-2) =-36 and x = 3, or, equivalently, z--36y + sl, x = 3. [Another method is
to use the vector (0,1,—36), parallel to the line, to form the parametric equations x = 3, y = 2+t,
z = 9 - 36/.1
42.34
State a set of conditions under which the mixed partial derivatives f xy (x 0 , y a ) and f yx (x 0 , y 0 ) are equal.
If (x 0 , y 0 ) is inside an open disk throughout which f xy and/^ exist, and if f xy andf yx are continuous at (jc 0 , y 0 ),
then f xy (x 0 , y 0 ) = f yx (x 0 , y 0 ). Similar conditions ensure equality for n>3 partial differentiations, regardless of the order in which the derivatives are taken.
42.35 For f(x, y) = 3x2y - 2xy + 5y2, verify that fxy=fyic.
f x = 6xy-2y, fxy=6x-2. fy =3x2-2x + Wy, fyx = 6x-2.
42.36 For f(x, y) = x7 In y + sinxy, verify fxy=fyx.