# Math Help - rate of change and tangent plane question

1. ## rate of change and tangent plane question

Exam practice question to which I don't have an answer. Would be greatful for any help.

Question.

Consider the function $f^2:R^3->R$ defined by

$f(x,y,z)=x^3+y^2-6xy+2z$,

and let P=(2,1,3).

(i) Find the rate of change of f(x,y,z) at the point P in the direction $v=(1,-1,1)^T$.

(ii) Write down the direction of maximum increase of f at P, and the rate of maximum increase. Show that P is on the surface $f(x,y,z)=3$ and write down the equation of the tangent plane to the surface at P.

(iii) There is another point Q on the surface $f(x,y,z)=3$ where the tangent plane is parallel to the tangent plane at P. Find Q.

(i) first, I find partial derivatives of f at P

$f_x=3x^2-6y=6, f_y=2y-6x=-10, f_z=2$

Let $\varphi$ be the angle between the above vector (6,-10,2)^T and the unit vector v=(1,-1,1)^T. Then I can find alpha:

$|\nabla{f}||v|cos\varphi=<\nabla{f},v>$

$cos\varphi=\frac{<\nabla{f},v>}{|\nabla{f}||v|}=\f rac{6+10+2}{\sqrt{36+100+4}\sqrt{3}}=\frac{18}{2\s qrt{105}}$

Then the rate of change of $\nabla{f}$ in the direction of v=(1,-1,0)^T is just the length of the projection of this vector on the direction of v, which can be calculated as

$|\nabla{f}_u|=|\nabla{f}|cos\varphi=\sqrt{140}\fra c{18}{2\sqrt{105}}=6\sqrt{3}$

(ii) the maximum rate of change of f at P is just the length of its gradient vector at this point, ie $\sqrt{140}$ as per (i).

Since coordinates of P satisfy the equation $x^3+y^2-6xy+2x=3$, it lies on the surface $f(x,y,z)=3$.

The tangent plane to the surface $f(x,y,z)=3$ at P (2,1,3) is

$0=f_x(x-2)+f_y(y-1)+f_z(z-3)$

$0=6(x-2)-10(y-1)+2(z-3)$

$6x-10y+z=5$ is the tangent plane to the surface f(x,y,z)=3 at P(2,1,3).

For a parallel tangent plane, I understand that I need to find a point that satisfies the equation f(x,y,z)=3 AND an equation of a plane parallel to 6x-10y+z=5. However, I don't know yet how to find it.

2. Originally Posted by Volga
Exam practice question to which I don't have an answer. Would be greatful for any help.

Question.

Consider the function $f^2:R^3->R$ defined by

$f(x,y,z)=x^3+y^2-6xy+2z$,

and let P=(2,1,3).

(i) Find the rate of change of f(x,y,z) at the point P in the direction $v=(1,-1,1)^T$.

(ii) Write down the direction of maximum increase of f at P, and the rate of maximum increase. Show that P is on the surface $f(x,y,z)=3$ and write down the equation of the tangent plane to the surface at P.

(iii) There is another point Q on the surface $f(x,y,z)=3$ where the tangent plane is parallel to the tangent plane at P. Find Q.

(i) first, I find partial derivatives of f at P

$f_x=3x^2-6y=6, f_y=2y-6x=-10, f_z=2$

Let $\varphi$ be the angle between the above vector (6,-10,2)^T and the unit vector v=(1,-1,1)^T. Then I can find alpha:

$|\nabla{f}||v|cos\varphi=<\nabla{f},v>$

$cos\varphi=\frac{<\nabla{f},v>}{|\nabla{f}||v|}=\f rac{6+10+2}{\sqrt{36+100+4}\sqrt{3}}=\frac{18}{2\s qrt{105}}$

Then the rate of change of $\nabla{f}$ in the direction of v=(1,-1,0)^T is just the length of the projection of this vector on the direction of v, which can be calculated as

$|\nabla{f}_u|=|\nabla{f}|cos\varphi=\sqrt{140}\fra c{18}{2\sqrt{105}}=6\sqrt{3}$
Not much point in using $\nabla f$ three different times is there? Simpler to remember that the derivative of f in the direction of unit vector u is just $\nabla f\cdot u$. The unit vector in the direction of <1, -1, 1> is $\left<1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}\right>$ so the derivative in that direction is
$\left<6, -10, 2\right>\cdot<1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}\right>= \frac{6+ 10+ 2}{\sqrt{3}}= \frac{18}{\sqrt{3}}= 6\sqrt{3}$ as you say.

(ii) the maximum rate of change of f at P is just the length of its gradient vector at this point, ie $\sqrt{140}$ as per (i).
Yes that is correct.

Since coordinates of P satisfy the equation $x^3+y^2-6xy+2x=3$, it lies on the surface $f(x,y,z)=3$.

The tangent plane to the surface $f(x,y,z)=3$ at P (2,1,3) is

$0=f_x(x-2)+f_y(y-1)+f_z(z-3)$

$0=6(x-2)-10(y-1)+2(z-3)$

$6x-10y+z=5$ is the tangent plane to the surface f(x,y,z)=3 at P(2,1,3).

For a parallel tangent plane, I understand that I need to find a point that satisfies the equation f(x,y,z)=3 AND an equation of a plane parallel to 6x-10y+z=5. However, I don't know yet how to find it.
You need to find another point on that surface so that $f_x= 3x^2- 6y= 6$, $f_y= 2y- 6x= -10$, and $f_z= 2$. The last is always true. Solve the first two for x and y. There will be two solutions, of course, one of which is x= 2, y= 1. Check that the other solution also satisfies f(x,y,z)= 3.

3. Thank you!!

So I solve $3x^2-6y=6, 2y-6x=-10$ and I get

$x_1=2, x_2=4; y_1=7, y_2=1$ ie the two points satisfying these equations are (2,1) and (4,7).

Now what do I do for z?... Should i use the fact that z satisfies $f_z=2$?

And, more importantly, how do I get closer to finding a tangent plane passing through this point that is parallel the one identified in (ii)?... Would that parallel plane has the same left side but different right side?

4. I shouldn't have said "Check that the other solution also satisfies f(x,y,z)= 3". I should have said "Find z so that f(4, 7, z)= 3".

5. Right. I think I got it now:

Given x=4, y=7 I can find z that satisfied the equation f(x,y,z)=3: z=-29. Therefore, the point Q (4,7,-29) lies on that surface.

Moreover, the normal vector to the surface at this point Q is (6,-10,2)^T, ie the same normal vector as normal vector to the same surface at the point P (2,1,3). Therefore, tangent planes at these two points (P and Q) are parallel.

I don't need to find the equation of the tangent plane at Q, but suppose I do: would it be

$0=f_x(x-4)+f_y(y-7)+f_z(z+29)$

$0=6(x-4)-10(y-7)+2(z+29)$, or

$6x-10y+2z=104$

Now, is there any connection between 5 (in the first tangent plane equation) and 104 here?...

6. It is essentially saying that the distance between those two parallel planes is 104- 5= 99.