I normally wouldn't do this, but I'm completely stuck. You don't have to explain the work that you're doing, but please show your steps so that I can follow along. If you can only do one or two of the problems, that's fine too.\n\n1.Consider the curve defined by x^2 + xy + y^2 = 27\n(a)Write an expression for the slope of the curve at any point (x, y).\n(b)Determine whether the lines tangent to the curve at the x-intercepts of the curve are parallel. Show the analysis that leads to your conclusion.\n(c)Find the points on the curve where the lines tangent to the curve are vertical.\n\n2.A circle is inscribed in a square as shown in the figure above. The circumference of the circle is increasing at a constant rate of 6 inches per second. As the circle expands, the square expands to maintain the condition of tangency. \n(a)Find the rate at which the perimeter of the square is increasing. Indicate units of measure.\n\nCont. in additional details...
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1 (a):  Bit of partial differentiation.  The slope is dy/dx; however, to get this we need to take partial derivatives of the function so we can calculate it.\n\ndy / dx = (Df / Dx) * (Dy / Df) [where the D denotes partial derivative].\n\nDf / Dx = 2x + y;\nDf / Dy = 2y + x.\n\nHence dy / dx = (2x + y) / (2y + x).\n\n1 (b):  x-intercepts of the curve have x = 0, and hence gradient is given by:\n\ndy / dx = (2*0 + y) / (2*y + 0) = y / 2y = 1/2.\n\nThis is the same for both intercepts, and so they are parallel.\n\n1 (c):  For vertical tangents, dy/dx is infinite; hence the denominator must be zero.  \n\n2y + x = 0   =>  x = -2y.\n\nSubstitute in:\n\n4y^2 - 2y^2 + y^2 = 27\n3y^2 = 27\ny^2 = 9\ny = +/- 3 and x = -/+ 6.\n\n----------------------------------------------------------\n\n2 (a):  Circumference = 2pi*r, and is changing by 6" / sec.\n\nHence d/dt [2pi*r] = 6 and dr / dt = 3/pi.\n\nSquare perimeter P is 8r, and so dP / dt is 8*3/pi = 24/pi inches / second.\n\n2 (b):  The area between the two is 4*r^2 - pi*r^2;\n\nA = (4 - pi)r^2.\ndA / dt = (4 - pi) * 2r * (dr / dt).\n\ndr / dt we have, from (a), and we know that the circle's area is 25pi:\n\n25pi = pi*r^2\nr^2 = 25\nr = 5\n\nSo:  dA / dt = 2*(4 - pi)* 5 * (3 / pi) inches squared / second.  And that might need a calculator :)
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For this problem,\n\nArea of rectangle = length*width.  The length is defined by 2x because the two bottom corners are and 'x' distance from the origin.  The width is determine by the curve 16-x^2.  This is the constraint we need to use.  In other words:\n\nArea = (2x)(y)\n\nSince y = 16-x^2, substitute:\n\nArea = (2x)(16-x^2) = 32x-2x^3\n\nTo maximize area, take the derivative and set equal to zero:\n\nA' = 32-6x^2 = 0, solving for x, you get = +4*sqrt(3)/3, -4*sqrt(3)/3.  We need to take positive numbers because we are dealing with area, which cannot be a negative number.\n\nNow plug this extreme value into the area.  We know that \n\nlength = 2x\nwidth = 16-x^2, now we can find the dimensions of the rectangle.\n\nLength = 8*sqrt(3)/3\nWidth = 16-(4*sqrt(3)/3)^2 = 16-(16/3) = 32/3\n\nThese are your dimensions\n---------------\n\nHope this helps
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uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
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1) Let sides of each square be x\nTherefore dimensions of box are (4 - 2x) by (3 - 2x) by x\nV =  x(4 - 2x)(3 - 2x)\n= 12x  - 14x² + 4x³\ndV/dx = 12 - 28x + 12x²\n= 4(3 - 7x + 3x²)\n= 0 for stationary points\nSo x = (7 ± √(49 - 36))/6\n= (7 ± √(13))/6\n= (7 + √(13))/6, (7 - √(13))/6\n\nd²V/dx² = -28 + 12x > 0 when x = (7 + √(13))/6\nand < 0 when x = (7 - √(13))/6\n\nTherefore maximum value when x = (7 - √(13))/6 ≈0.566 units.\nSo Vmax = (7 - √(13))/6 *(4 - 2*(7 - √(13))/6) * (3 - 2*(7 - √(13))/6)\n≈3.032 units³\n\n2) y = √(x + 1)\nLet P (t, √(t + 1)) be on the curve and let Q = (3, 0)\n\nThus |PQ|² = (3 - t)² + (0 - √(t + 1))²\n= 10 - 5t + t²\n\nIf |PQ| is minimum so is |PQ|²\n\nd|PQ|²/dt = -5 + 2t = 0 for stationary points and as this is a quadratic with a> 0 then it is a minimum\n\nSo t = 2.5\nWhence the required point is (2.5, √(3.5))
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QUESTION 1:\n\nAsk yourself: WHICH TRIG FUNCTION IS OPPOSITE OVER ADJACENT?  IT IS THE TANGENT FUNCTION, RIGHT?\n\nThe bottom of the tree is 25 feet away from a point that measures 30 degrees from ground level to the top of the tree.\n\nAnswer: tan30degrees = x/25feet...choice 1.\n\nNotice that they are NOT asking you to solve for x.  Get it?  \n\nQUESTION 2:\n\nYou need to write the given equation 3x - 2y = 12 in the y =mx + b\nform.\n\nLet's do that right now.\n\nWe have:\n\n3x - 2y = 12 \n\nI will subtract -3x from both sides of the equation.\n\nWe get:\n\n-2y = -3x + 12...See what I did?  \n\nNext:\n\nDivide BOTH sides of the equation by -2 to find the value of y.\n\nWe get this:\n\ny = (-3x + 12)/-2\n\nWe can reduce the equation to this:\n\ny = 3/2 (x) - 6\n\nSlope is 3/2 and the y-intercept is -6.\n\nFinal answer: choice 1\n\nGuido
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sigh...you shouldn't wait this late to do your homework...and you should study more, but i'll show you how to do this one - hopefully it'll help you with the other ones you posted lol.\n\nyou know the line is 2x-3y=18\n\nfind the slope of this line since you'll use it later:\n\n2x-3y=18\n-3y=18-2x\ny=-6+(2/3)x\n\nthus the slope is 2/3, since in that equation, y=b+mx, m=2/3.\n\nnow the line you want to find is perpendicular to this.  meaning the slope is the negative reciprocal.\n\nnew slope = -1/m = -1/(2/3) = -3/2\n\nnow you found the new m.  plug this into the given slope-interceot form equation and you get:\n\ny=mx+b\ny=-(3/2)x+b\n\nnow since you know that this line passes through point (-1,8), plug these y and x values into the equation and solve for b:\n\ny=-(3/2)x+b\n8=-(3/2)(-1)+b\n8=3/2+b\nb=8-3/2\nb=16/2-3/2\nb=13/2\n\nnow plug b back into your equation and you get:\n\ny=-(3/2)x+b\ny=-(3/2)x+13/2\n\nwhich is your answer :)  now go try and solve the rest of your problems by yourself.\n\n[[edit]]\n\nand don't listen to the guy above me - he's wrong and i don't like beer.
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Part a)\n\nx = width\ny = length\n2x + 2y = 1000, so y = 500 - x\n\nA = xy = x(500 - x) = 500x - x^2\n\nb)  x should be positive (and small enough to fit the plot of land that he is planting the garden in)\n\nc)  If one corner is at the coordinates (0,0), the opposite corner is at (x,y) as defined above.\n\nd)  A = 500x - x^2\n\nCalculus Solution:\n\ndA/dx = 500 - 2x\nWhen 500 - 2x = 0, the area is a maximum.  Here, x = 250 and y = 250, so the maximum area is 62,500 square yards.\n\nNon-calculus Solution:\n\nGraph the equation A = 500x - x^2.  You can see that the parabola crosses the x-axis at x = 0 and at x = 500.  By symmetry, you know that the peak of the parabola must be half-way between 0 and 500 at x = 250.  When x = 250, A = 62,500, which is the maximum area.
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- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
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1.) If you graph a "linear function" all you have is a line.  The rate of change is the slope.  The change in y is the "rise" and the change in x is the "run", and the slope is "rise/run", so in this case, the rate of change is 3/9, or 1/3\n\n2.) slope is, again, the change in y divided by the change in x.  Given 2 coordinates, the formula for slope is\n\n(y2 - y1)/(x2 - x1)\n\nSo in this case it's (-7 - (-4))/(3-2) which equals -3/1, or -3\n\n3.) If this means that the slope = 1, then the equation of a line is y = mx + b, and you need to solve for the y-intercept, b.  You know that the slope, m, equals 1, so\n\ny = 1x + b\n\nYou also have a coordinate, (3,8), so plug it into the equation\n\n8 = 1(3) + b\n8 = 3 + b\nb = 5\n\nSo the equation would be y = x + 5\n\n4.) The slope of a perpindicular line is the negative reciprocal of the original line.\n\nx - 3y = 4  solve for y\nx - 4 = 3y\ny = (1/3)x - 4/3\n\nSo the slope for this line is 1/3, therefore the slope for the perpindicular line is - 3\n\n5.) solve the equation for y to get it in the form of y = mx + b\n\n3x - 4y = 4\n3x - 4 = 4y\ny = (3/4)x - 4/3\ny = mx + b\n\nSlope is "m" and y-intercept is b, so slope is 3/4 and y-intercept is -4/3
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x^2=cost^2\n(y/2)^2=sint^2\n\nsint^2+cost^2=1\nso...\nx^2+y^2/4=1\n\nfind implicit differential\n2x+2y dy/dx =0 \ndy/dx= -x/y\nsubstitute the point (2/3, sq.rt.3) for x, y\nu get the slope \n\nthen the tangent line equation is\nx - 2/3 = slope (y- 3)
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a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
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area of circle is A = pie r^2.\ndifferentiating with respect to t, we get,\ndA/dt = pie * 2*r*( dr/dt)\nnow, we know, the circumference is 2*pie*r. let us take 2*pie*r as C.therefore,\n2*pie*r  = C   and \nrate of change of radius dr/dt is - 0.1 cm/sec....(note the minus as its decreasing)\ntherefore rate of change of area in terms of  circumference is\ndA/dt = - C * (0.1 cm/sec)\ntherefore, answer b is correct.
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1) y = ax + b, then a is the slope is tan of the angle with the x-axis\nx-3y=4 => y = (1/3) x - 4/3 , the atan(1/3) is the angle\n\n2)  x^3 + 2x^2 + 4x + 9\ntry divisiors of 9 , 1, -1, 3,-3 bnone of them work so there are no rational zeros. ( only irrationals )\n\n3) tan(x) = -1/2\ni dont do this one , -30 degrees, draw a rectangle and you have the solution
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Must have x + y > 0, else the expression is negative or ±∞. Multiply by x + y to get\n\n¼ ( x + y ) < x² + y² < ½ ( x + y )\n\nx² + y² - ¼ x - ¼y > 0  and x² + y² - ½ x - ½y < 0\n\n( x - 1/8 )² + ( y - 1/8 )² > 1 / 32 and ( x - ¼ )² + ( y - ¼ )² < 1/8\n\n( x - 1/8 )² + ( y - 1/8 )² > 1 / 32  are all points outside a circle centred at ( 1/8 , 1/8 ) with radius √2 / 8.\n\n ( x - ¼ )² + ( y - ¼ )² < 1/8 are all points inside the circle centred at ( ¼ , ¼ ) with radius √2 / 4.\n\nThe first circle is inside the second and touches it at (0,0).\n\nThe required condition is satisfied by all points outside the small circle and inside the large circle. Use QuickMath (Plot -> Advanced) to see the region.
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I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
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A short element of the curve is dL=sqrt(dr^2+(r*dt)^2), r*dt elementary length of the arc with radius r and polar angle t (your theta), dr elementary increase of r for dt.\nThus dL=sqrt(r’^2+r^2)*dt.\nIf  r=a*(1-cost), then dL=a*sqrt((sint)^2+(1-cost)^2)=a*sqrt(2)*sqrt(1-cost)*dt. \nFortunately 1-cost=2*(sin(t/2))^2 or dL=2a*sin(t/2)*dt, L=-4a*cos(t/2) for t=0 until 2pi, thus L=-4*a*(-1-1)=8*a – Oh! check it poor kid!\nIf r=(sin(t/2))^2=0.5*(1-cost), then a=0.5 for t=0 until pi. It must be done likewise. Thus L=-4*0.5*(0-1)=2
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The arc length of the curve y = f(x) from x = a to x = b is\n\nL = ∫[a to b]√{ 1 + [f '(x)]² } dx\n\nRewrite the curve as f(x) = 3 x^(4/3) - (3/16) x^(2/3) from x = -216 to 216. [I believe you made a typo with 3/12, else the following will get messy!]\n\nf '(x) = 4 x^(1/3) - (1/8) x^(-1/3)\n\nAfter simplifying, and using symmetry\n\nL = 2 ∫[0 to 216]√[ 16 x^(2/3) + (1/64) x^(-2/3) ] dx\n\nLet t = x^(2/3) and get\n\nL = 3 ∫[0 to 36]√( 16 t² + 1/64 ) dt\n\nLet u = 4t to get\n\nL = ¾ ∫[0 to 144]√( u² + 1/64 ) du\n\nThis evaluates to\n\n¾ [0 to 144 of \n½ u √( u² + 1/64 ) + ½ (1/64) ln | u + √( u² + 1/64 ) | ]\n\nwhich simplifies to\n\n54 √( 144² + 1/64 ) + \n( 3 / 512 ) { ln [ 144 + √( 144² + 1/64 ) ] - ln ( 1/8 )\n\n= (27/4) √1327105 + ( 3 / 512 ) ln ( 1152 + √1327105 )\n\n= 7776.048295\n\nVerify using WIMS Function calculator