[2] #1#2

Homework Six Please be aware that there are several typos, due to the conversion program which put the solutions onto the web. I am sorry, but at the moment, I cannot repair these.

Section 13.6 # 1. Using the chain rulle $\frac{dz}{dt} = \Pd{z}{x} \Pd{x}{t} + \Pd{z}{y} \Pd{y}{t}$. The appropriate partial derivatives are: $\Pd{z}{x} = y^2$, $\Pd{z}{y} = 2xy$, $\Pd{x}{t} = -te^{-t}$, and $\Pd{y}{t} = \cos t$. Thus, $\Pd{z}{t} = -te^{-t} y^2 + 2xy\cos t$. To get the final answer, insert the functions x(t) and y(t) to get $\Pd{z}{t} = e^{-t} \left[ -t\sin^2 t + 2 \sin t \cos t \right]$. #2. The chain rule gives the same general formula for $\Pd{z}{t}$ as in number 1. The appropriate derivatives are: $\Pd{z}{x} = \sin y + y\cos x$, $\Pd{z}{y} = x\cos y + \sin x$, $\Pd{x}{t} = 2t$, and $\Pd{y}{t} = \frac{1}{t}$. Thus, $\Pd{z}{t} = 2t (\sin y + y\cos x) + (1/t) (x\cos y + \sin x) = 2t \sin (\ln t) + t \cos (\ln t) + 2t \ln t \cos t^2 + (1/t) \sin t^2$. #7. The chain rule for z = f(x,y) with x=x(u,v) and y=y(u,v) gives us $\Pd{z}{u} = \Pd{z}{x} \Pd{x}{u} + \Pd{z}{y} \Pd{y}{u}$ and $\Pd{z}{v} = \Pd{z}{x} \Pd{x}{v} + \Pd{z}{y} \Pd{y}{v}$. The appropriate derivatives are $\Pd{z}{x} = e^{-y} - ye^{-x}$, $\Pd{z}{y} = -xe^{-y} + e^{-x}$, $\Pd{x}{u} = \sin v$, $\Pd{x}{v} = u\cos v$, $\Pd{y}{u} = -v \sin u$, and $\Pd{y}{v} = \cos u$. Thus, we find that the two partial derivatives of z are:

$$\Pd{z}{u} = \left[e^{-v\cos u} + v\cos u e^{-u\sin v}\right]\... ... \left[ -u\sin v e^{-v\cos u} + e^{-u\sin v} \right] v\sin u$$

$$\Pd{z}{v} = \left[e^{-v\cos u} + v\cos u e^{-u\sin v}\right]u... ... \left[ -u\sin v e^{-v\cos u} + e^{-u\sin v} \right] \cos u$$

#8. The appropriate derivatives are: $\Pd{z}{x} = -2x\sin(x^2+y^2)$, $\Pd{z}{y} = -2y\sin (x^2+y^2)$, $\Pd{x}{u} = \cos v$, $\Pd{x}{v} = -u\sin v$, $\Pd{y}{u} = \sin v$, and $\Pd{y}{v} = u\cos v$. Using the chain rule as written in problem 7, we find

$$\Pd{z}{u} = \sin(u^2) \left[ -2u\cos^2 v - 2u\sin^2 v \right] = -2u\sin(u^2)$$

$$\Pd{z}{v} = \sin(u^2) \left[ 2u^2\cos v \sin v -2u^2\sin v \cos v\right] = 0$$

#19. An example of a homogeneous function is the function f(x,y) = x² + y² since f(tx,ty) = (tx)² + (ty)² = t² x² + t² y² = t² (x² + y²) = t² f(x,y). Thus, for this function, p = 2. Other functions would potentially have different values of p. To simplify matters, let u = tx and v = ty. Then, for g(t) = f(tx, ty) = f(u,v) we have that $g'(t) = \Pd{f}{u} \Pd{u}{t} + \Pd{f}{v} \Pd{v}{t}$. Now, u_t = x and v_t = y. Also, at t = 1 we have that u=x and v=y so that g'(1) = xf_x + yf_y. This is the right hand side of the statement to be proved. However, since f(x,y) is a homogeneous function, we also have that g(t) = t^p f(x,y). Taking derivatives, we find that g'(t) = p t^p-1 f(x,y) so that g'(1) = p f(x,y). Since these two ways of computing g'(1) must be equal, the statement

x f_x(x,y) + y f_y(x,y) = p f(x,y)

is proved. Section 13.7 #3. For the function f(x,y) = xe^y:

$$f_x = e^y, \qquad f_y = xe^y$$

$$f_{xx} = \Pd{e^y}{x} = 0, \qquad f_{xy} = \Pd{e^y}{y} = e^y$$

$$f_{yx} = \Pd{xe^y}{x} = e^y, \qquad f_{yy} = \Pd{xe^y}{y} = xe^y$$

#8. For the function $f(x,y) \tan^{-1} (x+y)$ we use the fact that the derivative of the arctangent (the inverse tangent) is given by $\frac{d}{dx} (\mbox{arctan}(x)) = \frac{1}{1+x^2}$ and the chain rule to find that

$$\Pd{f}{x} = \frac{1}{1+(x+y)^2} \Pd{(x+y)}{x} = \frac{1}{1+(x+y)^2}, \qquad \Pd{f}{y} = \frac{1}{1+(x+y)^2}$$

$$f_{xx} = \Pd{}{x} \frac{1}{1+(x+y)^2} = \frac{-1}{[1+(x+y)^2]^2} \Pd{[1+(x+y)^2]}{x} = \frac{-2(x+y)}{[1+(x+y)^2]^2}$$

One quickly find that all of the second partial derivatives are equal. This is to be expected, since the oringinal function f depends on both x and y in the same way. #9. If z = f(x) + y g(x) then we see that z_y = g(x) and z_yy = 0 since the function depends only linearly on y. Thus, the cross sections of this function for fixed values of x will all be lines in a plane parallel to the yz plane. #10. If z_xy = 4y then (a) z_yx = 4y since the mixed partial derivatives of a continuous function are the same, (b) z_xyx = 0 since z_xyx = (4y)_x = 0, and (c) z_xyy = 4 since z_xyy = (4y)_y = 4. For problems #11 through #19, here are the answers as a list. For each problem, the signs of the derivatives are listed in the order f_x, f_y, f_xx, f, and f_xy.

#11. +, 0, +, 0, 0

#12. +, 0, -, 0, 0

#13. -, 0, -, 0, 0

#14. -, 0, +, 0, 0

#15. 0, -, 0, -, 0

#16. 0, +, 0, -, 0

#17. -, -, 0, 0, 0

#18. +, -, -, -, +

#19. -, +, +, +, -

Review of Integration For each of the following problems, the name of the method used to most easily obtain the antiderivative is given first. If you want practice, try just looking at these hints and not the final answer. If you're stuck, or sure you've got it right, check the answer.

#1. [U substitution] Let $u = \cos (3\theta)$ so that $du = -3\sin(3\theta) d\theta$ or $sin(3\theta) d\theta = -\frac{1}{3} du$ then the integral becomes

$$\int \sin(3\theta) \cos^2 (3\theta) d\theta = - \frac{1}{3} \... ...rac{1}{3} \frac{u^3}{3} + C = -\frac{\cos^3(3\theta)}{9} + C.$$

#2. [U substitution] Let u = 4^3t. Then (using the correct derivative law, which is NOT the power law) $du = (3) 4^{3t} \ln 4 dt$ or $4^{3t} dt = \frac{1}{3\ln 4} du$. The integral is then

$$\int 4^{3t} dt = \int \frac{1}{3\ln 4} du = \frac{u}{3\ln 4} + C = \frac{4^{3t}}{3\ln 4} + C.$$

#3. [Power rule-after simplification] Split the fraction up into three terms $\frac{x^4+x^2+1}{x^3} = x + \frac{1}{x} + x^{-3}$ then integrate with the normal power law to get

$$\int \frac{x^4+x^2+1}{x^3} dx = \int \left[ x + \frac{1}{x} +... ...} \right] dx = \frac{1}{2} x^2 + \ln x - \frac{1}{2x^2} + C.$$

#4. [U substitution] Try letting $u = \sqrt{x}$ so that $du = \frac{1}{\sqrt{x}} dx$ and then $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = \int e^u du = e^u + C$. The final answer is then $e^{\sqrt{x}} + C$.

#5. [Trig Substitution-or memorization] Let $x = \tan u$ so that $dx = \sec^2 u du$ the integral then becomes

$$\int \frac{1}{1+x^2} dx = \int \frac{1}{1+\tan^2 u} \sec^2 u du.$$

Now the trig identities tell us that $1 + \tan^2 u = \sec^2 u$ so that the integral is really just $\int du = u + C = \mbox{arctan} x + C$.

#6. [Exponential Rule-after simplification] Since $\frac{1}{e^m} = e^{-m}$ we can simply let u = -m so that du = -dm and the integral is

$$\int e^{-m} dm = - \int e^u du = -e^u + C = - e^{-m} + C.$$

#7. [U substitution] Let u = 4v + 7 then du = 4 dv and

$$\int \sqrt[3]{4v+7} dv = \int (4v+7)^{1/3} dv = \int \frac{1}... ...} \frac{3}{4} u^{4/3} + C = \frac{3}{16} (4v + 7)^{4/3} + C.$$

#8. [Power Rule-after simplification] First multiply out the polynomial so that you get (1 + 3r²)² = 1 + 6r² + 9r⁴ and then integrate to get $\int (1 + 3r^2)^2 dr = r + 2r^3 + \frac{9}{5}r^5 + C$.

#9. [U substitution] Let $u = 3 - \sec \alpha$ so that $du = - \sec \alpha \tan \alpha d\alpha$. Then the integral becomes $\int \frac{1}{u} du = ln u + C$. So that the final answer is $ln (3-\sec \alpha) + C$.

#10. [U substitution-after simplification] Use the properties of logarithms to write $\ln (y^2) = 2 \ln (y)$ then let $u = \ln y$. This makes $du = \frac{dy}{y}$ which is already present in the integral. Thus, the integral is really $\int 2 u du = u^2 + C = (\ln y)^2 + C$.

#11. [U substitution] Let u = w³ so that du = 3w² dw or $\frac{1}{3} du = w^2 dw$. The integral is then $\int \frac{1}{3} \sin u du = -\frac{1}{3}\cos (w^3) + C$.

#12. [Simplification] Try writing the integrand as $\tan^2 \theta = \frac{\sin^2\theta}{\cos^2\theta}$. The numerator of this can be rewritten as $1 - \cos^2 \theta$ so that the integrand is really $\frac{1}{\cos^2 \theta} - 1 = \sec^2 \theta - 1$. The integral is then easily computed as $\int (\sec^2 \theta - 1) d\theta = \tan \theta - \theta + C$.

#13. [Rewrite the Integrand] Recall that $\frac{1}{\cos t}$ is just $\sec t$ so the integrand is really $\sec^2 t$. Recall also that $\frac{d}{dt} \tan t = \sec^2 t$ so the integral is just $\tan t + C$.

#14. [U Substitution-with Simplification] The denominator of the integrand can be factored to the form (x+3)². Now a simple u substitution of u = x + 3 results in the integration

$$\int \frac{1}{x^2 + 6x + 9} dx = \int \frac{1}{(x+3)^2} dx = \int u^{-2} du = -u^{-1} + C = -\frac{1}{x+3} + C.$$