Math1001 Linear Algebra tut5_solution

T HE U NIVERSITY OF S YDNEYS CHOOL OF M ATHEMATICS AND S TATISTICSSolutions to Tutorial5Web Page:.au/u/UG/SS/SS1001/Preparatory questions1.For each function,find the indicated derivative(ordinary or partial).Recall that to computef x(x,y),regard y as a constant and differentiate f(x,y)with respect to x.Similarly,to computef y(x,y)regard x as a constant and differentiate f(x,y)with respect to y.(a)f(x)=3x2−16;d f∂x and∂fdy Solution:f′(y)=15e3y(d)f(x,y)=xe3y;∂f∂ySolution:f x(x,y)=e3y,f y(x,y)=3xe3y (e)f(x)=4ln x;d fx(f)f(x,y)=y ln x;∂f∂ySolution:f x(x,y)=yx2+4;d f(x2+4)2=8−2x2x2+y2;∂f∂ySolution:By the quotient rule,f x(x,y)=y(y2−x2)(x2+y2)2. Copyright c 2011The University of Sydney12.Let f(x,y)=x3+x2y3−2y2.Calculate the partial derivatives f x(x,y)and f y(x,y)andfind thevalue of each derivative at the point(1,2).Solution:By direct calculation:f x(x,y)=3x2+2xy3f x(1,2)=19,f y(x,y)=3x2y2−4y f y(1,2)=4.Questions to do in class3.Let f(x,y)=2x−3y+2.(a)Find the equation of the tangent plane to the surface z=f(x,y)at the point(x,y)=(3,1).Solution:Since f(x,y)=2x−3y+2,we have f x(x,y)=2and f y(x,y)=−3.In par-ticular,f x(3,1)=2and f y(3,1)=−3.Therefore,the equation of the tangent plane toz=f(x,y)when(x,y)=(3,1)isz−5=2(x−3)−3(y−1),which can be rearranged into either of the equivalent forms,z=2x−3y+2,2x−3y−z=−2.(b)Can you explain the relationship between the tangent plane and the surface z=f(x,y)?Solution:The tangent plane is exactly the original surface z=f(x,y).This is what weshould expect because,just as the tangent line to a line at any point is the same line,thetangent plane to a plane at any point is the same plane.4.Find thefirst-order partial derivatives f x(x,y)and f y(x,y)of the following functions:(a)f(x,y)=√2√x+√2√2√∂x =2x,∂z∂x (2,1)=4,∂zTherefore,the equation of tangent plane isz−8=4(x−2)+8(y−1),which can be rearranged toz=4(x+2y−2).6.Recall the“informal”definition of the limit of a function f(x)as x approaches the number c.We say lim x→c f(x)=ℓif we can make f(x)as close as we like toℓfor all x sufficientlyclose to(but not equal to)c.For each of the functions given by the formulas below,sketch the graph and decide whether or not the function has a limit as x→1.(Note:Only an intuitive idea of the limit of a function is required here.Limits and limit laws will be discussed in more detail later in the course.) (a)f(x)= 1if x<1,−x if x>1.Solution:If there is a limit,it must be a single numberℓwith the property that values of f(x)can be made as close as we like toℓby choosing x sufficiently close to1.The graph shows that when x is less than1,f(x)=1and so if the limit exists it should be1.However,when x is arbitrarily close to1 but greater than1,the values of f(x)are approaching−1and so the limit,if it exists,should be −1.Thus it is impossible to have a singleℓwhich is compatible with the informal definition;the function does not have a limit as x→1.(b)f(x)= x if x<1,2−x if x>1.Solution:For this function the limit exists and is equal to1.We can make f(x)as close to1as we like by choosing x sufficiently close to1(but not equal to1).(Although not required for the answer,the3following sample calculations may help you to understand the process.If we want,for example, f(x)to be within0.01of1(i.e.0.99<f(x)<1.01)then we must choose x to be no further away from1than0.01units,i.e.0.99<x<1or1<x<1.01.If we want f(x)to be within0.001of1 (i.e.0.999<f(x)<1.001)then we must choose x to be no further away from1than0.001units, i.e.0.999<x<1or1<x<1.001.)(c)f(x)= x2−1if x<1,√∂V ∂V∂P=−1.Solution:Since P=kT∂V=−kTP,we have∂VP.SinceT=PV∂P=V∂V∂V∂P= −kT P V V P=−1.Questions for further practice8.Find the ordinary derivatives of the following functions.(a)f(x)=sin xe2x =cos x−sin x(b)g(y)=tan(y2+cos y)Solution:The chain rule gives g′(y)=(2y−sin y)sec2(y2+cos y) (c)h(w)=ln(w+sin(w3+4))Solution:The chain rule gives h′(w)=1+3w2cos(w3+4)t2+1Solution:The quotient rule givesF′(t)=(t2+1)(3t2+10t)−(t3+5t2−1)(2t)(t2+1)2(e)G(u)=21+cos vSolution:The quotient rule gives H′(v)=(1+cos v)(2tan v sec2v)+tan2v sin v∂x ,∂f2x+3y;f y(2,4)Solution:f y(x,y)=32x+3y,f y(2,4)=34+12=3∂y(1,0)Solution:∂f∂y(1,0)=f y(1,0)=−1+3=2.(d)z=x√√∂x,∂z∂x=√2x−3/2 =√2x3/2,∂z2y−1/2 −1x=x y−1x.(e)f(u,v)=tan−1 u∂u,∂f1+(u/v)21u2+v2,f v(u,v)=1v2 =−u10.Find the twofirst-order partial derivatives of the following functions:(a)w=cos u sin vSolution:∂w∂v=cos u cos v.(b)V=πr2aSolution:∂V∂a=πr2.11.Find the equation of the tangent plane to the given surface at the specified point.(a)z=5+(x−1)2+(y+2)2;(2,0,10).Solution:∂z∂y=2(y+2)=4, when(x,y)=(2,0).Therefore,the equation of the tangent plane isz−10=2(x−2)+4(y−0),which can be rearranged toz=2(x+2y+3).(b)z=sin(x+y);(1,−1,0).Solution:We have∂z∂y=cos(x+y)=cos0=1,when(x,y)=(1,−1).Therefore,the equation of the tangent plane isz−0=1(x−1)+1(y+1),which can be rearranged toz=x+y.(c)z=ln(2x+y);(−1,3,0).Solution:We have∂z2x+y =2,∂z2x+y=1,when(x,y)=(−1,3).Therefore,the equation of the tangent plane isz−0=2(x+1)+1(y−3),which can be rearranged toz=2x+y−1.12.(Harder)Show that the ellipsoid3x2+2y2+z2=9and the sphere x2+y2+z2−8x−6y−8z+24=0are tangential to each other at(1,1,2).That is,show that these two surfaces have a common tangent plane at the point(1,1,2).Solution:The ellipsoid is centred at the origin and has semiaxes a=√2and c=3.The sphere has centre(4,3,4)and radius r=√9−3x2−2y2.6Now,rewrite the given equation of the sphere as(x −4)2+(y −3)2+(z −4)2=17.We then see that the point (1,1,2)lies on the lower half of the sphere (since the sphere is centred at (4,3,4).Solving for z and taking the negative square root (why?)gives the equation,z =k (x ,y )=4−2,h y =−2y 9−3x 2−2y 2,h y (1,1)=−1.Hence,the equation of the tangent plane is z −2=−(3/2)(x −1)−(y −1),which rearranges to3x +2y +2z =9.Similarly,the equation of the tangent plane to the surface z =k (x ,y )at (1,1)isz −2=k x (1,1)(x −1)+k y (1,1)(y −1).The required derivatives arek x =x −417−(x −4)2−(y −3)2,k x (1,1)=−3。