Practice for Final Exam MAT 240

Practice for final exam MAT 240

(1) For the vectors u = < 6, 3, 2> and v = < 2, -2, 0>:

(a) find 2u – 3v

(b) Find the angle between the two vectors

(2) Sketch the function, and graph the velocity and acceleration vectors at t = 1.

r(t) = ti + j

(3) Find the slope of the function in the x-direction and y-direction at (3,4).

(4) Find the gradient of the function and the maximum value of the dierctional derivative at .

(5) Give the tangential and normal components of acceleration for a particle moving along the curve:

r(t) = t2i + tj , at t = 1.

(6) A horizontal conveyor discharges gravel that falls into a truck bed 6 feet below and 4 feet horizontally away from the end of the conveyor. Determine the speed at which the conveyor belt should be moving so that the gravel falls onto the center of the truck bed.

Position Function for a Projectile -

(7) Find the minimum distance from the point (5,5,0) and the cone below:

(8) Find the area of the surface given by z= f(x,y) over the region R:

(9) Find the work done by the conservative force field

on an object moving along the path from (0,0) to (4,8).

(10) Find the volume of the solid:

z = 3 – x – y

Solutions:

(1)(a) 2< 6, 3, 2 > - 3< 2, -2, 0> = <12 – 6, 6 - - 6, 4 – 0> = < 6, 12, 4 >

(b)

(2) r(t) = ti + j, v = r'(t) = i + 2tj, a = r''(t) = 2j,

so at t = 1: v(1) = i + 2j, and a(1) = 2j (it's constant for all t).

(3)

(4) . The relevant formula is . Please note that the gradient is a vector-valued function, unlike the directional derivative, which is just a real number (when evaluated at a point).

Here, we have

I repeat: the output of an (evaluated) gradient is a vector, while the output of an (evaluated) directional derivative is a number.

(5)) r(t) = t2i + tj , at t = 1.

, so stop cryin', and find v, ||v|| and a by differentiating:

So I'm going to find the normal component by using the formula , so we need the ||a|| = , and then just let it rip with the formula:

(6)

gravel

conveyor 4 ft

In the model, the speed will just be the initial velocity of the gravel. Note that the initial height will be h0 = 6 feet and the launch angle will be , since the conveyor is horizontal. Think of the truck bed as height 0. So the question could be rephrased: What does the initial velocity v0 have to be so that the length (the gravel travels) is 4 + M when the height is 0? (when the gravel hits the bed). So:

(a) find out when the height is 0. Set

Now use this to solve for v0, by setting the length equal to M + 4:

(7) This is an optimization problem. The distance between any two points in space is given by , which is a function of three variables, which we need to get down to two variables in order to find the critical number. Plug in the point (5,5,0) first:

Now (i)the square root function is minimized or maximized depending on the radicand. In other words, we can simply ignore the square root sign and work on what's inside it, and (ii) we can substitute to get the function down to 2 variables:

and find the critical point by setting the partials equal to 0:

The critical point is

the minimum distance would be the z-coordinate there:

(8) We use polars as the base region is round – the circle of radius 4 centered at the origin. R:

Now, simplify the radical:

The surface area is given by this integral evaluated using the formula I derived in class:

(9) We are given that the field is conservative, so we:

(i) find the potential function:

Union the terms:

. This is our potential function. By independence of path, we can plug

(ii) plug the endpoints into f:

Note that we essentially ignored the given path . That's what independence of path means – if the field is conservative, it doesn't matter which route you take from endpoint to endpoint: the work done will be the same.

(10) (2) Give the base region R as vertically simple:

Then set up the double integral: