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Section 6.6 Tangent Lines and Implicit Differentiation

These answers correspond to the problems in Section 2.4.

  1. (a) \(f^\prime(1)=-7\text{;}\) (b) The slope equals to \(-7\text{;}\) (c) \(y=-7x+3\text{.}\)

  2. (a) See Figure 6.6.1.(b) Observe that all lines through \((3,1)\) are given by \(y=1+k(x-3)\) and then find \(k\) so that the equation \(17-2x^2=1+k(x-3)\) has a unique solution. \(k=-16\) or \(k=-8\text{.}\) (c) \(y=-16x+49\) and \(y=-8x+25\text{.}\)

    Figure 6.6.1. \(y=17-2x^2\) and its two tangents through \((3,1)\)

  3. \(3\text{.}\) The question is to find \(a\in\mathbb{R}^+\) such that the circle \(x^2+(y-a)^2=1\) and the parabola \(y=x^2\) have the same tangent lines at their intersection points. \(\ds a=\frac{5}{4}\text{.}\)

  4. (b) \(1\text{.}\) (c) \(y=3x+5\)

  5. Solve \(\ds y'=\cosh x=1\text{.}\) The point is \((0,0)\text{.}\)

  6. Solutions of \(-a^3=3a^2(4-a)\) are \(a=0\) and \(a=6\text{.}\) The points are \((0,0)\) and \((6,216)\text{.}\)

  7. \(y=-4x-5\text{.}\)

  8. \(y=2\sqrt{3}x\pm 3\text{.}\)

  9. (a) \(\ds -\frac{\pi }{4}\text{,}\) (b) \(\ds y =\sqrt{2}x +1-\frac{\pi }{4}\text{.}\)

  10. \(y=x-1\text{.}\)

  11. \(y=e^4(4x-7)\text{.}\)

  12. \(c=\pm6\text{.}\)

  13. \(a= 2\pm\sqrt{3}\text{.}\)

  14. Yes. Use the Intermediate Value Theorem for the function \(y^\prime(x)\text{.}\) Alternatively, solve \(y^\prime=20\text{.}\)

  15. We note that \(y^\prime=3(x-1)^2\text{.}\) Two lines, none of them horizontal, are perpendicular to each other if the product of their slopes equals \(-1\text{.}\) Thus to find all points on the curve \(C\) with the property that the tangent line is perpendicular to the line \(L\) we solve the equation \(\ds -\frac{1}{3}\cdot 3(x-1)^2=-1\text{.}\) Hence \(x=0\) or \(x=2\text{.}\) The lines are \(y=3x-1\) and \(y=3x-5\text{.}\) See Figure 6.6.2.

    Figure 6.6.2. \(y=(x-1)^3\) and \(3y+x=0\)

  16. \(h^\prime(0)= 2\text{.}\)

  17. From \(\ds e^y\cdot \left( \frac{dy}{dx}\cdot \ln (x+y)+\frac{1+\frac{dy}{dx}}{x+y}\right) =-\left( y+\frac{dy}{dx}\right) \cdot \sin (xy)\) it follows that \(\ds \left. \frac{dy}{dx}\right| _{x=1}=-1\text{.}\)

  18. \(\ds \frac{dy}{dx}=\frac{y-x^4}{y^4-x}\text{.}\)

  19. \(\ds y^\prime=\frac{-3x^2 y-2x+y^2}{x^3-2xy}\text{.}\)

  20. \(\ds y\ln x =x\ln y\\ \frac{dy}{dx}\cdot \ln x+\frac{y}{x}=\ln y+\frac{x}{y}\cdot \frac{dy}{dx}\\ \frac{dy}{dx}=\frac{y(x\ln y-y)}{x(y\ln x -x)}\text{.}\)

  21. \(\ds \frac{dy}{dx}=\frac{3^x\ln 3 +\sinh y}{e^y-x\cosh y}\text{.}\)

  22. \(\ds y^\prime=\frac{2y^2 e^xy}{1-2xye^xy-2e^xy}\text{.}\)

  23. \(\ds \frac{dy}{dx}=\frac{\cosh x -2xy}{x^2-\sin y}\text{.}\)

  24. \(\ds \frac{dy}{dx}=\frac{1-y(x-y)}{1+(x-y)(x+3y^2)}\text{.}\)

  25. \(\ds y^\prime = \frac{y(2x+\cos (y)-1}{xy\sin(y)+1}\text{.}\)

  26. \(\ds y^\prime = \frac{y\sin(x)-\sin (y)}{x\cos(y)+\cos? (x)}\text{.}\)

  27. \(\ds y^\prime = \frac{2x+\cos (y)-1}{x\sin(y)+\cos (y)}\text{.}\)

  28. \(\ds y^\prime = \frac{(y^2+1)(2x+\sin? (x))}{e^y y^2+e^y-1}\text{.}\)

  29. \(\ds y^\prime = \frac{2xy\sin(x^2 )+\sin(y^2)}{\cos (x^2 )-2xy\cos(y^2)}\text{.}\)

  30. \(\ds y^\prime = \frac{(2x^2-1)(y^2+1)}{x(y^2 \sec^2 (y)+\tan^2 (y))}\text{.}\)

  31. \(\ds y^\prime = \frac{\sec^2 (x+y^3 )-2y(2xy^ +1)}{3y^2 \sec^2 (x+y^3 )-2x(xy+1)}\text{.}\)

  32. (a) \(x+y=0\text{;}\) (b) The graph crosses the \(x\)-axis at the points \((\pm \sqrt{3},0)\text{.}\) The claim follows from the fact that \(2x-y-xy'+2yy'=0\) implies that if \(x=\pm \sqrt{3}\) and \(y=0\) then \(y'=2\text{.}\)

  33. \(\ds \frac{10}{21}\text{.}\)

  34. \(x+y=\pi\text{.}\)

  35. \(y=0\text{.}\)

  36. \(\ds y^\prime(3)=\frac{10}{21}\text{.}\)

  37. \(\ds y-\pi=\frac{1}{3}(x-\pi)\text{.}\)

  38. \(\ds \frac{4}{3}\text{.}\)

  39. \(x=0\text{,}\) \(x=2\sqrt{3}{4}\text{.}\)

  40. (a) \(\ds \frac{dy}{dx}=-\frac{2xy}{x^2+2ay}\text{;}\) (b) We solve the system of equations \(1+a=b\text{,}\) \(\ds -\frac{2}{1+2a}=-\frac{4}{3}\) to get \(\ds a=\frac{1}{4}\) and \(\ds b=\frac{5}{4}\text{.}\)

  41. From \(\ds \frac{dy}{dx}=-\sqrt{\frac{y}{x}}\) we get that the tangent line \(l\) to the curve at any of its points \((a,b)\) is given by \(y-b=-\sqrt{\frac{b}{a}}(x-a)\text{.}\) The sum of the \(x\)-intercept and the \(y\)-intercept of \(l\) is given by \((a+\sqrt{ab})+(b+\sqrt{ab})=(\sqrt{a}+\sqrt{b})^2=k\text{.}\)

  42. From \(\ds \frac{2}{3\sqrt[3]{x}}+\frac{2y'}{3\sqrt[3]{y}}=0\) we conclude that \(\ds y'=-\sqrt[3]{\frac{y}{x}}\text{.}\) Thus the tangent line through the point \((a,b)\) on the curve is given by \(\ds y-b=-\sqrt[3]{\frac{b}{a}}(x-a)\text{.}\) Its \(x\) and \(y\) intercepts are \(\ds \left( a+\sqrt[3]{ab^2},0\right)\) and \(\ds \left( 0,b+\sqrt[3]{a^2b}\right)\text{.}\) Thus the square of the portion of the tangent line cut off by the coordinate axis is \(\ds \left( a+\sqrt[3]{ab^2}\right) ^2+\left( b+\sqrt[3]{a^2b}\right) ^2=a^2+2a\sqrt[3]{ab^2}+ b\sqrt[3]{a^2b} + b^2+2b\sqrt[3]{a^2b}+a\sqrt[3]{ab^2}=\left( \sqrt[3]{a^2}+\sqrt[3]{b^2}\right) ^3=9^3\text{.}\) The length of the portion is \(\sqrt{9^3}=27\text{.}\)

  43. \(\ds y+4=\frac{3}{4}(x-8)\text{.}\)

  44. (a) \((0,0)\text{,}\) \((0,\pm 2)\text{.}\) (b) \(\ds y^\prime =\frac{x(2x^2-5)}{2y(y^2-2)}\text{.}\) (c) \(x=\sqrt{5}\text{.}\)

  45. (a) \(y=3x-9\text{;}\) (b) \(y(2.98)\approx 3\cdot 2.98-9=-0.06\)

  46. (a) \(y'(4)=4\text{,}\) \(y"(4)=-11\text{;}\) (b) \(y(3.95)\approx -0.2\text{;}\) (c) Since the curve is concave down, the tangent line is above the curve and the approximation is an overestimate.