## Review for final exam (MA241 Spring 2013)

1. Find the angle between the vectors $v=\langle 2,0,1 \rangle$ and $w=\langle 4,-2,0 \rangle.$
2. Find the curvature at $t=0$ of the curve $\boldsymbol{r}(t) = \langle \cos(2t), 4t, \sin(2t) \rangle.$
3. Find parametric equations for the tangent line at the point $(5,0,2)$ to the curve with parametric equations
$\begin{cases} x(t) =3+2\sqrt{t} \\ y(t)=t^3-t\\ z(t) = t^3+t \end{cases}$
4. Find the length of the curve $\boldsymbol{r}(t) = 6t \boldsymbol{i} + 8t^{3/2} \boldsymbol{j} + 6t^2 \boldsymbol{k},$ for $0 \leq t \leq 1.$
5. Find an equation of the plane that contains the point $(1,3,5)$ and is perpendicular to the line with equation $\big\{ x=3+t, y=3t, z=5-2t \big\}.$
6. Find an equation of the plane that contains the line $\big\{ x=3+4t, y=2, z=4t \big\}$ parallel to the plane $3x+6y-3z=18.$
7. Find the distance from the point $(3,7,-5)$ to each of the following:
• The $xy-$plane
• The $y-$axis.
8. Find the volume of a parallelepiped with adjacent edges $PQ, PR, PS,$ where $P=(3,0,3), Q=(-1,3,6), R=(5,3,1), S=(2,4,4).$
9. At what points does the helix $\boldsymbol{r}(t)= \langle \sin(t), \cos(t), t \rangle$ intersects the sphere $x^2+y^2+z^2=10?$
10. Find a non-zero vector orthogonal to the plane through the points $P=(-2,0,0), Q=(0,3,0), R=(0,0,1).$
11. Compute the partial derivatives $\partial z/\partial x$ and $\partial z/\partial y$ if $x, y$ and $z$ are related by the equation $yz = \ln (x+z).$
12. Find a linear approximation to the function $f(x,y) = \sqrt{17-x^2-4y^2}$ at the point $(1,0).$ Use it to approximate the value of $f(1.92,0.91).$
13. Find an equation of the tangent plane to the surface of $z=5x^2-y^2+5y^2$ at the point $(-1,5,5).$
14. Use the chain rule to compute $dz/dt$ if $z=x^2+y^2+xy, x=\sin t, y=e^t.$
15. Find the directional derivative at $(2,1)$ in the direction of vector $\boldsymbol{v} = 10 \boldsymbol{i} + 5\boldsymbol{j}$ for the function $g(r,s) = \tan^{-1}(rs).$
16. For the function $f(x,y)=\sqrt{64-4x^2-4y^2},$
• Find domain an range
• Sketch the level curves for $k=0,1,4.$
17. Use differentials to estimate the amount of metal in a closed cylindrical can that is 26 cm high and 6 cm in diameter, if the metal in the top and the bottom is 0.3 cm thick, and the metal in the side is 0.05 cm thick.
18. Find the maximum rate of change of $f(x,y,z)= \displaystyle{\frac{x+5y}{z}}$ at the point $(4,1,-1).$
19. Find all second partial derivatives of the function $f(x,y)=\sin^2(mx+ny).$
20. Find all the extreme values of $f(x,y)=x^2+2y^2$ on the disk $x^2+y^2 \leq 1.$
21. Find all maxima, minima and saddle points of the function $f(x,y)=2x^3+xy^2+5x^2+y^2+8.$
22. Find the absolute maximum and absolute minimum values of $f(x,y)=4x+6y-x^2-y^2+7$ on the set $D= \big\{ (x,y) : 0 \leq x \leq 4, 0 \leq y \leq 5 \big\}.$
23. Find the mass and center of mass of the lamina with density $\rho(x,y)=xy^2$ that occupies the region $D=\big\{ (x,y) : 0 \leq x \leq 2, -1\leq y \leq 1 \big\}.$

24. Evaluate $\displaystyle{\iiint_E xyz\, dV},$ where $E$ is the region that lies between the spheres $\rho=2, \rho=4$, and above the cone $\varphi=\pi/3.$

25. Evaluate $\displaystyle{\iint_D 4xy-y^2\, dA},$ with domain $D$ being the region bounded by $y=\sqrt{x}$ and $y=x^3.$

26. Evaluate the triple integral $\displaystyle{\iiint_E 2x\, dV},$ where $E$ is the solid bounded by the paraboloid $z=2x^2+2y^2,$ above the plane $z=8,$ and between the planes $y=0$ and $y=2.$

27. Is the vector field $\boldsymbol{F}(x,y) = \langle ye^x+\sin x\cos y, e^x + \cos x \sin y \rangle$ conservative? If it is, find a function $f(x,y)$ so that $\boldsymbol{F} =\nabla f.$

28. Evaluate the triple integral $\displaystyle{ \iiint_E \frac{1}{\sqrt{y^2+z^2+1}}\, dV},$ where $E$ is bounded by the paraboloid $y^2+z^2=1$ and the plane $x=1.$

29. Evaluate the integral $\displaystyle{\oint_C (3y^2-e^x)\, dx + (7x+\sin y)\, dy},$ where $C$ is the circle $x^2+y^2=4.$

30. Prove that the Jacobian to change from Cartesian to Spherical coordinates is $\rho^2 \sin \varphi$.

31. Find the area of the region $R$ bounded by the circles $x^2+y^2=16, x^2+y^2-4$ in the half-plane $x\geq 0.$

32. Evaluate the line integral $\displaystyle{ \int_C \big( \tfrac{x}{3} + 7y \big)\, ds},$ where $C$ is given in parametric equations by $x=3t, y=t, z=1+5t, (0 \leq t \leq 1).$
1. April 27, 2013 at 10:00 pm

I know this is probably a super simple question but I am spacing on the beginning material from the semester–for number 5, how do you calculate a perpendicular vector? Do you use the cross product? And if so, what’s the second vector you use?

2. April 27, 2013 at 10:14 pm

Also for number 6, if the plane contains the line, can I just use the vector from the plane and the point from the line and write the equation that way? Or does the vector in the line have a factor in the equation of the plane as well?

• April 28, 2013 at 3:13 pm

For this one recognize that you can use the normal vector given by 3x+6y-3z=18 to form the equation for the other plane because the planes are parallel. so the normal vector for one plane will also be a normal vector to the other plane.

• April 28, 2013 at 3:20 pm

Yes, I think you use the vector from the plane since it is parallel and the point from the line.

• April 28, 2013 at 9:49 pm

This problem is very similar to HW 2, #13 if you are still having problems on how to go about it.

3. April 27, 2013 at 11:02 pm

For number 11, do we have to find an equation in which z is in terms of x and y, and if so, how does that work with the ln?

• April 28, 2013 at 3:14 pm

For this one you use implicit differentiation.

• April 28, 2013 at 4:44 pm

No, you don’t need to find an equation in which z is in terms of x &y. You need to solve for \partial z/\partial x by taking the partial derivative of x in both sides and by keeping in mind that z is a function of x & y.

• April 28, 2013 at 9:20 pm

So what would be the partial in terms of x of ln(x+z)? Because z isn’t in terms of x and y at that point. Same with yz. I don’t understand how you can take the partial derivative of a function without knowing that function in the variables x and y.

• April 28, 2013 at 11:23 pm

implicit differentiation is hard for me to explain. this link may help. Its example 4 near the bottom of the page http://tutorial.math.lamar.edu/Classes/CalcIII/PartialDerivatives.aspx

• April 29, 2013 at 12:09 am

To find dz/dx: I first got y(dz/dx)= (1+(dz/dx))/(x+z); then I got the (dz/dx) terms to one side which leads you to:
(dz/dx)((x+z)y-1)=1; which leads to the final answer
dz/dx = 1/((x+z)y-1)

For dz/dy just do the same thing: I got
dz/dy = (z(x+z))/(1-y(x+z))

• April 29, 2013 at 9:13 am

Ahhhhhhhh thank you. Makes way more sense.

• April 29, 2013 at 12:54 am

For dz/dx, you differentiate implicitly with respect to x. You treat y as a constant. So for the left side yz, you would get y*dz/dx. Am I on the right track? I implicitly differentiated both sides and then solved for dz/dx.

4. April 27, 2013 at 11:19 pm

For number 13 is one of the y^2 terms supposed to be just a y?

• April 29, 2013 at 12:15 am

No this is correct. Like usual just find f(sub x) and f(sub y). Then plug in the values of x and y into those equations. I got fx(-1,5)= -10; fy(-1,5)=40. Then just plug these values into the equation of the tangent plane. So its: z-5 = -10(x+1) + 40(y-5)… simplify, solve for z and you get the answer.

• April 29, 2013 at 7:00 pm

I think you are right, one of y^2 terms should be y because then the point (-1,5,5) is not on the plane and we can’t find the tangent line equation of a point that is not in the plane.

5. April 28, 2013 at 12:32 am

For 16 I am getting that the domain lies outside of the circle–does this work? Or does the domain need to be enclosed meaning I did something wrong?

• April 28, 2013 at 3:19 pm

I think it should be enclosed in this problem. Try it again setting Sqrt(64-x^2-y^2) >= 0 and get 4x^2+4y^2 by itself on one side then 64 on the other. if the x^2 + y^2 term is <= the term on the other side then the domain is going to lie within the circle.

• April 28, 2013 at 9:23 pm

You’re right–makes way more sense. My inequality was backwards from the beginning.

• April 28, 2013 at 9:40 pm

For the range–do I plug in the lower and upper bound for the domain and solve from there or just say [0, infinity) because that’s the capability of z?

• April 28, 2013 at 11:05 pm

The range is all the values that z can be for any value of x and y. The low end of your range [0 is right but z cannot go to infinity in this case.

• April 29, 2013 at 12:21 am

You are right about the minimum of 0 limit but the range maximum is not infinity. When you found the domain, you should have got the equation x^2+y^2<=16; therefore the max value of x or y is the square root of 16=4. Range = [0,4]

• April 29, 2013 at 8:02 pm

Wouldn’t the range here be [0,8] since the lowest point we have in the domain is (0,0). Whenever we plug in (0,0) in the function we get f(x,y)=sqrt(64)=8

• April 29, 2013 at 8:05 pm

For the level curves, did anyone get circles with r=4 for k=0 , r=sqrt(63)/2 for k=1 , and r=sqrt(15) for k=4 ?

• April 30, 2013 at 8:51 pm

You are right, I believe. The range is [0,8]. what did you get for your equation for the level curves? I got r=sqrt(16-(k^2/4))

6. April 28, 2013 at 4:38 pm

For # 12, do we need a specific point to find the linear approximation and then plug in the given point?

• April 28, 2013 at 7:39 pm

Yeah i think you would have to be provided a point

• April 28, 2013 at 9:16 pm

The specific point is (1.92, .91) and that goes in the (Xo, Yo) portion of the equation and the plain X and Y stay as variables. You should get an equation with a constant, an X term and a Y term.

• April 29, 2013 at 12:25 am

No, actually you do need a point. You can’t find a linear approximation without one. The equation is not a linear approximation.

• April 29, 2013 at 1:12 am

So do you take the partial derivatives and then plug the approximate value of x into the partial derivative of x, and approx. value of y into the partial derivative of y to find your answers?

• April 29, 2013 at 1:18 am

This is throwing me off also because in all of the examples in the book, you are given an indicated point to find the linear approximation at, then you use the approximation to find the value of say f(1.92,0.91). Can someone please explain this?

• April 29, 2013 at 7:02 pm

Prof. Blanco-Silva said it is at the point (1,0) today in class.

• April 29, 2013 at 8:08 pm

I got my final answer to be .23, can anyone confirm this?

• April 29, 2013 at 9:29 pm

I got my answer to be 3.77. But I might be wrong!
Did you get f(1,0)=16 , fx(1,0)=-1/4 , and fy(1,0)=0 ?
L(x,y)=(-1/4)x+(17/4)?

• April 30, 2013 at 3:12 pm

whats the formula you are using? Because I have the same values as you do for f(1,0) fx(1,0) and fy(1,0)

• April 30, 2013 at 11:09 pm

Collin, I have used L(x,y)=4+(-1/4)(x-1)+0(y-0)

7. April 28, 2013 at 9:08 pm

for #25 should the first integral be from x^3 to sqrt(x) or from sqrt(x) to x^3? I eventually came up with a negative answer so I’m doing something wrong. Also the second part goes from 0 to 1 right (where it converges)?

• April 28, 2013 at 10:54 pm

it would be from x^3 to sqrt(x) because x^3<=y<=sqrt(x)

• April 29, 2013 at 12:30 am

If you graph it: you can see that sqrt(x) is above x^3 from 0 to 1; so sqrt(x) should be on top. All i did was a double integral from {0 to 1 dx {x^3 to sqrt(x) dy of the terms 4xy-y^2. I got 19/60 but not sure if right.

8. April 28, 2013 at 9:26 pm

For number 26 should the question read ‘below the plane z=8’ instead of above?

• April 30, 2013 at 12:41 pm

Yes!

9. April 28, 2013 at 10:06 pm

Did anyone else get (19.5pi)cm^3 for number 17?

• April 28, 2013 at 10:42 pm

You should be getting around 33 cm^3

• April 29, 2013 at 12:34 am

I got (12.5pi)cm^3, V=(pi)(r^2)h is the equation for a cylinder.
Getting the differential you get
Delta V = 2pi(r)dr(h) +pi(r^2)(dh)
for dr= .05; dh =0.3; h=26; r=3

• April 29, 2013 at 9:24 am

Thanks, Jan–for some reason I was trying to use x, y, and z to solve it.

• April 29, 2013 at 9:37 am

Also just wanted to let you know I got (10.5pi)cm^3–I did the calculations a couple times and derived the same equation you did. Is this a mistake on my part?

• April 29, 2013 at 11:52 am

Yea, my bad. It is 10.5pi.

• April 30, 2013 at 8:09 pm

dh should be equal to .6 because the thickness change is on the top and bottom making the answer 13.2pi

10. April 28, 2013 at 10:30 pm

For 20 when I set the gradients equal to each other, I get 2=1…but 2 does not equal 1 and that’s a dead end for finding extreme values. Am I off in left field?

• April 29, 2013 at 12:43 am

You have to find the critical points. After finding the partial derivatives, you should set them equal together 2x=4y; thus one critical point is (0,0). From the equation of the disk; you can obtain 4 more critical points (0,+-1) and (+-1,0). Plugging these values into the original equation: max=2; min=0

• April 29, 2013 at 11:16 pm

So in this problem we can’t use Lagrange multiplier?!

11. April 29, 2013 at 9:52 am

For 21, I got that the minimum is (0,0) and (-8/5, 4/5) is a saddle point. Is this a common answer? Everything made sense in my calculations, but I’m slightly confused as to the properties–do you have to have a max?

• April 29, 2013 at 9:15 pm

I found there to be 2 local minimums: (0,0) and (-5/3,0). I found there to be 2 saddle points as well as (-1,2) and (-1,-2). I also did not find a maxima. Since my answers are different than yours I can run through how I got there and then compare. I began by first finding fsubx and fsuby and set both equal to 0 and found the first critical point at (0,0). While setting fsuby equla to zero I also found x to possible be x=-1. I tested this in and found two critical points at (-1,-2) and (-1,2) after plugging both x=-1 and y=0 into fsubx. The setting y=0 is fsubx, I found another critical point at (-5/3,0) which may be where me and you differ. Then finalized by using the second derivative test to find 2 local minima and 2 saddle points.

• April 29, 2013 at 11:56 pm

I found the same critical points as Collin, but I had different conclusions! (-1,2) is a saddle because Hess(-1,2) 0 and second partial derivative 0 and second partial derivative > 0

• April 29, 2013 at 11:58 pm

(-1,2) saddle , (0,0) min , (-1,-2) & (-5/3,0) are max

12. April 29, 2013 at 6:15 pm

For #18, once you get the magnitude of the gradient at the point, do you have to divide your new vector by the unit vector?

• April 29, 2013 at 6:17 pm

Nevermind, i made a careless error. I am pretty sure you divide your gradient by the magnitude of your gradient to give you the maximum rate of change. Somebody correct me if I am wrong

• April 29, 2013 at 6:56 pm

I think the maximum rate of change is just the gradient, but from what I understand, it’s just an accepted practice to turn it into its unit vector form.

• April 29, 2013 at 8:02 pm

Oh, ok. Thanks, that makes sense.

• April 29, 2013 at 8:58 pm

I think the maximum rate of change is the magnitude of the gradient according to theorem 15 in page 916. The gradient at the point ( after you divide it be the magnitude) is just the direction where the maximum change occur. Please correct me if I’m wrong.

I found the maximum rate of change to be sqrt(198). Did y’all get the same answer?

13. April 29, 2013 at 7:06 pm

In # 15, do we need to change the vector to a unit vector? if yes, did anyone get 3/sqrt(125) as answer?

• April 29, 2013 at 7:40 pm

and if we want to find the direction of where the max change occur, Is it the vector given in the problem or the maximum occurs in the unit vector of the gradient?

• April 29, 2013 at 8:13 pm

For #15, I found the gradient and the unit vector first. Then I computed the dot product of the gradient and the unit vector to give me the directional derivative. I ended up with 4/sqrt(125), but I may have made a mistake somewhere that I have not found yet.

For the gradient, I got and for the unit vector I got .

• April 29, 2013 at 8:14 pm

EDIT: gradient = (1/5,2/5)
U = (10/sqrt(125), 5/sqrt(125))

• April 29, 2013 at 9:06 pm

I got the same thing. Thanks. and for my second question I think the maximum change happens in (1,2) which is the unit vector of (1/5,2/5) ( this is not required for question # 15 )

14. April 29, 2013 at 8:43 pm

For number 17, I saw many of you guys came up with an answer at 10.5pi. Going back through the equation yall were using however, when you found dh to be .3, you didn’t factor in that you need to multiply it by 2 since it is on the top and bottom. So therefore instead of .3=dh, it would be .6=dh. Is this correct or do you only factor in the .3 once?

• April 29, 2013 at 9:17 pm

You are absolutely right. it has to be multiplied by 2 because it is in the top and bottom and I got 12.3 pi. You can see the solution for similar problem in webassign HW # 7 problem # 17.
Why do we chose dh=.3 and dr=.05 and not the opposite ?

15. April 29, 2013 at 8:51 pm

For #24, I missed part of this on the last exam. If E lies above the cone phi = pi/3, then what are the boundaries of phi when you compute the triple integral? If someone could explain how they got this, I would really appreciate it.

• April 30, 2013 at 12:21 pm

Phi in spherical method starts at the z-axis where Phi=0 and doesn’t go further than Pi. In this problem since we are at the region above the cone we have 0<=phi<=pi/3 because our Phi can't go further than Pi/3.

• April 30, 2013 at 6:10 pm

Thanks, that cleared it up for me.

• April 30, 2013 at 8:00 pm

Did you all end up with the integral with 0<=theta<=2pi sin(theta)cos(theta) which ends up being 0??

16. April 29, 2013 at 9:02 pm

For #18 I know this may be a simple question but I found the Gradient Vector, then evaluated it at each point. Is the maximum rate of change just the magnitude of it?

• April 29, 2013 at 9:41 pm

Yes I think so. I got sqrt(198). did you get the same thing?

• April 29, 2013 at 10:17 pm

I got sqrt(97) as the maximum rate of change. You need to start out with the gradient of f, then plug in the point (which is (4, 1, -1) in this problem). Afterwards, do |gradient f(4, 1, -1)| (| = absolute value mark) and you should get your answer for max rate of change.

• April 29, 2013 at 11:22 pm

Thanks. I probably have gotten the wrong gradient then. Can you please tell me what you got for the gradient?

• April 29, 2013 at 11:26 pm

I got =

• April 30, 2013 at 3:15 pm

I got the sqrt(107) but I think I am messing up on taking the derivative of x,y,z,. When you take the derivative of fx do you treat z as a constant or is the answer for fx 1/z?

• April 30, 2013 at 7:03 pm

I got the sqrt(107) as well.
fx is 1/z and z is treated as a constant because you are differentiating with respect to x.

17. April 29, 2013 at 9:32 pm

For 22, I found absolute maximum to be 20 at (2,3) and an absolute minimum to be -32 at (8,-1). Can anyone confirm this or show me what they did instead if you got a different answer? Im skeptical of mine.

• April 29, 2013 at 9:35 pm

Nevermind just saw that I used the bounds given in problem 23 on accident.

• April 30, 2013 at 12:13 am

I got max of 20 at (2,3) and min of 7 at (0,0) and (4,0). Did you get the same thing?

• April 30, 2013 at 3:19 pm

Im still getting different answers. My critical points are (2,3) (0,4) (4,-4) (6,0) (-4,5). Can you go through how you got your answer?

18. April 29, 2013 at 9:40 pm

For # 19, did you get (partial f / partial x)= 2sin(mx+ny)(m*cos(mx+ny)) and then we need to use the product rule to find all second partial derivatives?

• April 30, 2013 at 3:24 pm

correct. I found fxx to be 2m^2cos(2mx+2ny), fyy to be 2n^2cos(2mx+2ny), and fxy to be 2mncos(2mx+2ny)

• April 30, 2013 at 3:28 pm

I found the same answer for the first derivative, and (partial f/ partial y)=2nsin(mx+ny)cos(mx+ny). I am not sure about the product rule because mine is coming up as a mess and I know there is a mistake in it but have not found it yet?

19. April 29, 2013 at 11:29 pm

Is #31 supposed to be solved through green’s theorem? If so is the question missing a detail? How would I set up a problem such as this?

Or is it a simple double integral?

• April 30, 2013 at 2:24 pm

I think it should be solved by using double integral.

• April 30, 2013 at 7:22 pm

If we need to use a double integral then what are the bounds?

• April 30, 2013 at 8:06 pm

I know the bounds of theta and rho for the double integral but what do I put underneath the integral? I’m drawing a blank

• April 30, 2013 at 8:07 pm

Is it just x^2 + y^2 converted to polar coordinates?

20. April 30, 2013 at 12:57 pm

In # 26 , do we use polar coordinates ?

• April 30, 2013 at 3:46 pm

Because its a triple integral and a paraboloid is involved, I think you would use cylindrical coordinates.

• April 30, 2013 at 6:15 pm

Hey, if you don’t mind, can you elaborate more on what you did on #26?

• April 30, 2013 at 8:04 pm

What limits did you use?

21. April 30, 2013 at 3:26 pm

What bounds did yall get for 24?

• April 30, 2013 at 6:40 pm

I got 2 <= p <= 4, 0 =< theta =< 2pi, 0 =< phi =< pi/3.

• April 30, 2013 at 6:49 pm

2 < p < 4
0 < phi < pi/3
0 < theta < 2pi

22. April 30, 2013 at 6:04 pm

I was hoping to be able to get some help on a few questions I am iffy about. Can anyone post their answers for 3, 9, 15, and 27?

• April 30, 2013 at 8:56 pm

3) x=5+t;y=2t;z=t^3+t
9)It was on homework 3 # 9
15) I am not sure but I got 4/(5*sqrt(5))

• April 30, 2013 at 8:58 pm

On number 3; I got z=2+4t.

23. April 30, 2013 at 6:35 pm

for 3 I got x=5+t, y=2t, z=2+4t
for 9 (sin3, cos3, 9) and (sin(-3),cos(-3), 9)
for 15 i got 5
and for 27i got ye^x-cos(x)cos(y)

These could be wrong, but thats what I have

24. April 30, 2013 at 8:03 pm

For 31 I know it’s supposed to be a double integral but what do I put under the double integral? I know 0<=theta<=pi and 2<=r<=4 but what's my equation?

• April 30, 2013 at 11:17 pm

Since we are trying to find the area, we don’t need to have an equation to integrate. We just integrate 1 and if you use polar which I used, you will integrate rho. Or without integrating you can just find the area of half the circles and find the difference between them. 1/2 pi r^2.

25. April 30, 2013 at 8:30 pm

Is #26 a Type II triple integral?

• April 30, 2013 at 8:32 pm

The reason I ask is because I am having trouble finding the x boundaries. This one has me stumped.

26. April 30, 2013 at 8:43 pm

Shouldnt problem 28 have another boundary for x?

• April 30, 2013 at 11:14 pm

No, because it is actually a cylinder not a paraboloid.

27. April 30, 2013 at 8:55 pm

Any one getting 4*sqrt(35) as answer for #32

• April 30, 2013 at 11:13 pm

Yes, I got the same answer.

28. April 30, 2013 at 11:21 pm

Can someone tell me what are the boundaries used for # 26. and is the answer = 0. ?

• April 30, 2013 at 11:30 pm

Ive been having trouble with it as well. What boundaries did you use? I used for y 0 to 2, x 0 to sqrt(4-y), and for z(iffy about) I have 2x^2+2y^2 to 8.

29. April 30, 2013 at 11:33 pm

Can someone explain their way through number 9?

1. No trackbacks yet.