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Homework Statement Integral of (sec(10x)^2)*(tan(10x)^6)dx Homework Equations The Attempt at a Solution The powers are throwing me off a little bit. I realize that the derivative of tan is sec^2, bt how will that help me with this problem?

Homework Statement ∫4to5 ∫1to2 (1x + y)−2 dy dx The Attempt at a Solution I am confused about what to do with this negative 2. Any ideas?

Duh. Thank you!

Homework Statement Consider the function f(x,y) = 4-x^2+3y^2 + y. Let S be the surface described by the equation z= f(x,y) where f(x,y) is given above. Find an equation for the plane tangent to S at the point (-1,0,3) The Attempt at a Solution Ok, SO i solved for the gradient of F...

Ok, I understand. Than you cepheid and Mark44 for helping me visualizing this problem! Thanks again! Loppyfoot

any ideas? Basically, I'm stuck on the problem: For what values of t (if any) exist when the vector <1,2t,3t^2> is parallel to <0,2,6t>. I know <0,2,6t>= k <1,2t,3t^2>. What do I do next?

COmputationally I feel like i am not getting anywhere. A=kv So , <0,2,6t> = k<1,2t,3t2> I get 0=k ; 2=2kt ; 6t=3t2k. Where should I go from here? Should I solve for k, because I tried that and I am going in circles...

So I found out how to solve to change the direction, not the speed, by doing the dot product between vectors v and a. How would I find the points when I solve for when they're parallel If v is parallel to a, there must be a constant k that links the two vectors being parallel. so a=kv...

Any Ideas?

Oh good point. thanks!

Homework Statement Find the velocity and acceleration vectors for the following curves. Determine the points where the velocity vector is changing DIRECTION but not changing SPEED, and the points where the velocity vector is changing SPEED but not DIRECTION. For: a. f(t) = (t,t2,t3)...

Alright! So the final answer after plugging in t=2 to get the slope, and f(2) to get the points, the parametric equation would be: f(t)= (4,1/4) + t<4,-1/4> Thanks a lot TIny TIm!

Homework Statement The function f(t)= (t2,1/t) represents a curve in the plane parametrically. Write an equation in parametric form for the tangent line to this curve at the point where t=2 The Attempt at a Solution I can solve the gradient from an implicit equation, but solving...

So since the gradient is <6,7>, the length of the gradient will be (36+49)^.5. So the length of the gradient is 9.22, but since it should be pointing in the direction of greatest decrease, the length of the gradient should be -9.22. correct?

Oh Ok. So let's say v=1, then w= -1. So the vector would be <1,-1>. So using that vector <1,-1> I'm then confused about how to solve the direction he should be headed towards? I shouldn't do the dot product because that would produce a scalar.. hm..

I'm having trouble visualizing how to find the vector (v,w)...

Ok so the gradient vector is: <2xy,x^2-2xy>, so once evaluated at point a, I get <6,7> for gradf(3,1). Should I do anything with a unit vector of (3,1), to find the directional derivative?

Homework Statement The equation T(x,y) = x2y-y2 + 180 is a temperature equation at point (x,y). If One is standing at point (3,1), what direction should he move in order to decrease the temperature? Secondly, if one moves a unit distance in this direction found above, how much should...

I'm having trouble with number 2. So I understand that for level sets, I look at the outputs and set them equal to some k. In my case, k=0. So, (t,t^2-cos(t))= 0. Should the first step be to solve for t? If so, how would I create an implicit equation from this problem? Thanks!

So the equation would be f(t)= (t,t,t^2-cos(t)), correct? Just making sure...

Ok Thank you HallsofIvy. So for another example for number one would be: If I have f(u,v) from R2-R3, = (u^2+v,u+v,u+v^2), The graph of this function would be: (u,v, U^2+v,u+v, u+v^2). Great, I understand now! Thanks a lot sir.

Multivariable Calc Image-Graph Problem! Homework Statement Given a function f: R-R2 , by f(t) = (t, t^2 - cos(t)), which represents a curve in the xy plane parametrically, give a function whose GRAPH represents this same curve. 2) Also, give a function h whose level set for height k=0...

Is that the process that should be done?

Is there a way to find that point computationally without sketching? Well, could I use the points (0,0,0), (1,1,0) and (2,0,2)? I got (2,0,2) by f(1).

Ok, so since the vector is pointing in the same direction as the line, I understand that I should use the cross product between two vectors. I understand that I have one vector, <1,-1,2> but how would I find the second vector? Should I use the vector <-1,1,-2> and take the cross product...

Homework Statement Find the implicit form for the plane that contains the origin and the line: L(t) = <1+t,1-t,2t> The Attempt at a Solution' So, the point P = (1,1,0) and vector v= <1,-1,2>. To find the implicit equation for the plane using the form ax + by + cz = d , I will...

any ideas?

Ok, Now I understand. So I should solve for k first: 2x= k ; 2y=0 ; -2z=2k . So x = k/2 and z=-k. If I substitute those back into the original equation, I get k^2/4+ y^2-k^2 = -1. But I have two variables. Should I have set the y=0 so I can solve for k?

So <2x,2y,-2z> dotted with <1,0,2> = 0 Ok, So if I do the dot product, I get: 2x-4z = 0 . So , x=2z. If I plug that into the original equation, I get: 4z^2 + y^2 -z^2 = -1 . Should I solve for y in terms of z?

Yes I understand that point. I'm confused on how to computationally solve this problem. Any thoughts?

Homework Statement Find all the points on the surface x^2+y^2-z^2=-1 where the tangent plane is perpendicular to the vector <1,0,2>. I'm confused! The Attempt at a Solution So, the gradient would be: <2x,2y,-2z>, but what do I do with the gradient vector to find these points...

I got the answer: <-2,1> and <3,6>

Homework Statement Write the vector <1,7> as a sum of two vectors, one parallel to <2,-1> and one perpendicular to <2,-1> Homework Equations DOt Product The Attempt at a Solution I'm confused on where to begin this problem. Should I be using the dot product? Thanks

Homework Statement Consider the ellipsoid 4x^2+2y^2+z^2 = 19. Find all the points where the tangent plane to this ellipsoid is parallel to the plane 2y−8x+z = 0. Homework Equations The Attempt at a Solution So I found the normal vector to the tangent, <8x,4y,2z>. I also...

Homework Statement Let f(x,y) = 5y^(2)-(2x^(2)+xy) Then an implicit equation for the tangent plane to the graph of f at the point (0,-2) is Homework Equations The Attempt at a Solution I understand that I should take the derivative to find the gradient vector. For the...

so the equation of the tangent line is y=-2x-3. How would I translate the y=mx+b into the parametric form?

So the slope of the tangent line would be: y'=4x+2...plug in x=-1. slope of tangent line at x=-1 is y'=-2. A point on the line would be (-1,-1). How would I translate this data into parametric form

Homework Statement The planes 3y-4x-4z = -18 and 3x-2y+3z = 14 are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is: L(t)= ? Homework Equations Cross Product Seems like it would be relevant here, but how...

Homework Statement The parametric form for the tangent line to the graph of y = 2x^(2)+2x-1 at x = -1 is Homework Equations The Attempt at a Solution I am confused about where to begin this problem. Any thoughts? Thanks!

No I have not. Is there a way to solve these computationally?

Homework Statement Consider the line L(t) = <1-3t,0>. Then: L is parallel, perpendicular or neither, to the Line L(t)= <6t-2,-3> Homework Equations The Attempt at a Solution I am stumped!

Homework Statement Consider the line L(t) =<2-t,1+4t,4+2t> and the point P =(5,0,-4). How far is P from the line L? Homework Equations The Attempt at a Solution I'm confused on how to being this problem. Any ideas would be great!

So, first I realized that v1= K<2,5,-4> Then I figured that since v1+v2=sum, ; then: <4,1,1> - K<2,5,-4> = v2 Now, what process would I use to find K?

Homework Statement Find two vectors, v1, and v2, whose sum is <-4,1,1>, where v1 is parallel to <2,5,-4>, and where v2 is perpendicular to <2,5,-4> Homework Equations I am guessing I use the cross product for this equation, but I'm confused about how to start this problem. The...

Ok, so should I set: 3t2-t3= 0? Then solve for t?

Would it be coming off of it at a right angle? Could I use the Pythagorean Theorem?

Homework Statement For what value(s) of t is <3t2-t3, 2t2> a vertical vector? The Attempt at a Solution I set them equal to each other, and received t=1. Is that what I am supposed to do?

Does Anyone have any Ideas? I don't think its that difficult of a question for some, but its just a little bit confusing.

Homework Statement The diagram above represents a collection of level sets for a certain function, where the outer-most level is at the lowest height. What are points A-E? relative min, relative max, saddle point, or not a critical point The Attempt at a Solution I have tried...

Anyone available to help?