Mock final exam Math - fall 7 Fernando Guevara Vasquez December 5 7. Consider the curve r(t) = ti + tj + 5 t t k, t. (a) Show that the curve lies on a sphere centered at the origin. (b) Where does the tangent line at t = interesects the yz plane? (a) We simply need to show that the distance r(t) to the origin is constant. Thus we write, r(t) = t + t + 5 t t = 5 which means that r(t) = 5, or equivalently that the curve lies in the sphere of radius 5 centered at the origin. (b) The tangent line at t passes through t and has direction vector r (t ), therefore its parametric equation is T(t) = r(t ) + (t t )r (t ). Plugging in the expression for r(t) we get r() = (, 5, ). Moreover we have, ( ) r (t) =, t, t, 5 t t thus evaluating at t =, r () = (, 5/, 7), and the tangent line at t = t has the parametric equation, + (t ) T(t) = 5 + 5 (t ). 7(t ) Now we note that points in the yz plane satisfy the equation x =, thus to find the intersection we must have that the x component of T(t) is zero. This means t =, and thus the intersection of the tangent line to the yz plane is T() = (, 5, 5).
. Sketch the level curve of f(x, y) = x/y that goes through the point p = (4, ). Calculate f(p) and draw the vector 4 f(p), placing its initial point at p. What should be true about f(p)? The level curves are the curves satisfying f(x, y) = k. Here the level curves are of the form x = ky (parabolas). For p = (4, ), we have f(p) =, therefore the level curve that passes through p is given by the equation x = y (see figure ). The gradient f(p) = (/4, ) should be perpendicular to the (tangent of the) level curve at p as can be seen in figure. Figure Level curve of f(x, y) = x/y passing through (4, ). (4, ) x = y y 4 f (4, ) 4 x 9. A particle moves according to r(t) = (t + )i + t j. Find the tangential and normal components of the acceleration at t. The velocity is v = r (t) = (, t) and the acceleration a = r (t) = (, ). The acceleration in the basis of the unit tangential T and unit normal N vectors is a = a T T + a N N. () We recall that the tangential vector is the unit vector with same direction as the velocity, that is T = r / r. To find a T we dot both sides of () by T and use that T T = and N T =. Thus, a T = a T = (, ) (, t) = 4. + 4t + 4t
Now that we know a T, we can find a N very easily using the Pythagorean theorem: a N = 4. Find the gradient vector of a a T = 4 6t + 4t = + 4t. F (x, y, z) = x + y + z + at p = (,, ). Write the equation of the tangent plane to the surface F (x, y, z) = 6 at p. We compute the gradient F (p ) = (x, 4y, z), which evaluated at p gives F (p ) = (, 4, ). The point p belongs to the surface since F (p ) = 6. We known that the tangent plane to the surface F (x, y, z) = 6 goes through p and is orthogonal to the gradient F (p ). Therefore its equation is given by F (p ) (p p ) =, which in the case of p = (,, ) gives (x ) + 4(y ) + (z ) =. Note: I we had z = f(x, y) instead, this is covered by what we did above. Indeed take F (x, y, z) =, where F (x, y, z) = f(x, y, z) z. Then the equation of the tangent plane at (x, y, z ) with z = f(x, y ) is given by, ( ) f x x y y =. z z This translates into the familiar equation of a tangent plane (that can be recovered using Taylor s theorem): z = f(x, y ) + f(x, y ) (x x, y y ). 5. Find the minimum distance between the point (,, ) and the surface z = + x + y. Points belonging to the surface are of the form (x, y, + x + y ). The distance (squared, which is recommended because otherwise the distance function is not differentiable at the origin) from such a point to (,, ) is given by (x, y, + x + y ) (,, ) = (x ) + (y ) + + x + y = x + y x y + =: f(x, y).
Thus we need to solve the unconstrained optimization problem min f(x, y). This can be achieved by first finding the critical points, which are all stationary points for f, that is points (x, y) for which f(x, y) =. We get that f(x, y) = (4x, 4y ), and the stationary point (/, /). To check if this is a minimizer we use the second partials test. We compute, D(x, y) = f xx f xy = 4 4 = 6 >. f xy f yy This means that the critical point at hand is not a saddle point. To check whether it is a minimum or a maximum we look at the sign of f xx. In our case f xx (/, /) = 4 >, therefore the point (/, /) is a minimizer of f(x, y). The minimal squared distance is f(/, /) = + =. 6. Assume we can measure the height and radius of a cylinder to within % accuracy. Estimate using differentials the maximum percentage error of the calculated volume. The volume of such a cylinder is V (r, y) = πr h. The change in volume V = V (r + r, h + h) V (r, h) due to the measurement errors r = ±.r and h = ±.h can be estimated using differentials as follows, V dv = V V r + r h h = πrh(±r.) + πr (±h.) πr h. + πr h. =.πr h, where the inequality in the chain is the triangle inequality. Thus we get that the relative error in the volume of the cylinder V /V is approximatively %. 7. Evaluate the mass of a ring R of height, inner radius and outer radius, where the mass density is proportional to the distance from the axis of symmetry. As can be seen from the quick sketch in figure, this integration begs for using cylindrical coordinates. The distance from the axis of symmetry is r, thus we take m(r, θ, z) = kr, where k > is some constant. We compute the mass as follows, π M = m(x, y, z)dxdydz = drkr dz dθ R ( = k ) ()(π) = 4kπ. 4
Figure The ring of problem 7. z y x 8. Calculate the area A(S) of the region S delimited by y = x and y = x in the following ways: (a) directly using double integrals (b) using the identity derived from Green s theorem, A(S) = xdy, where S has been oriented counter-clockwise as usual. See figure for the region of interest and for how we partitioned the boundary S of the region into two smooth paths C and C. (a) Using double integrals we get that A(S) = x dx dy = x S dx(x x ) = = 6. (b) Using the identity involving line integrals we get that, xdy = xdy + xdy S C C = ydy + ydy = y/ + y = = 6. 5
Figure The region S of interest in problem 8. The boundary is oriented counter-clockwise, and divided into C (y = x, between x = and x = ) and C (y = x, between x = and x = ). y C y = x S C y = x x 6