Sign in to jmaths to print this page or save it as a PDF: jmaths-8ce638778fd6.herokuapp.com/maths/ai-hl/further-derivatives/ahl-2-2/problems/b/

jmaths

IB Mathematics: Applications and Interpretation HL · AHL 2: Further Derivatives · AHL 2.2

Tangents, normals and gradient problems Problem set B Worked solutions

Skills

  1. Differentiate with respect to x.

    1. (a) \(y=9e^x\)

      • The derivative of \(e^x\) is \(e^x\)
      • The constant stays: \(\frac{dy}{dx}=9e^x\)

    2. (b) \(y=-8e^x\)

      • \(\frac{dy}{dx}=-8e^x\)

    3. (c) \(y=-\dfrac{e^x}{6}\)

      • \(y=-\frac{e^x}{6}\) is \(-\frac16e^x\)
      • \(\frac{dy}{dx}=-\frac16e^x\)

    4. (d) \(y=\dfrac{7e^x}{3}\)

      • \(\frac{dy}{dx}=\frac73e^x\)

  2. Find the gradient of each curve at the given value of x. Give exact answers.

    1. (a) \(y=\sin x\) at \(x=\dfrac{\pi}{4}\).

      • \(y'=\cos x\)
      • At \(x=\frac\pi4\): \(\cos\frac\pi4=\frac{\sqrt2}{2}\)

    2. (b) \(y=2e^x+5x\) at \(x=\ln 6\).

      • \(y'=2e^x+5\)
      • At \(x=\ln6\): \(2\times6+5=17\)

  3. Find the point on the curve \(y=1+6x-2e^x\) where the gradient is 2. Give exact coordinates, then 3 significant figures.

    • \(y'=6-2e^x=2\), so \(e^x=2\) and \(x=\ln2\)
    • \(y=1+6\ln2-2\times2=6\ln2-3\)
    • Point \((\ln2,\,6\ln2-3)=(0.693,\,1.16)\)

  4. Find the equation of the tangent to the curve \(y=x^2-8x+3\) that is parallel to the line \(y=2x+7\).

    • \(y'=2x-8=2\), so \(x=5\)
    • \(y=25-40+3=-12\): the point is \((5,-12)\)
    • \(y+12=2(x-5)\), so \(y=2x-22\)

  5. Find the equation of each tangent.

    1. (a) \(y=6\sin x\), \(0<x<\dfrac\pi2\): the tangent parallel to \(y-3x=5\).

      • Parallel to \(y-3x=5\) (gradient 3): \(y'=6\cos x=3\), so \(\cos x=\frac12\) and \(x=\frac\pi3\)
      • \(y=6\sin\frac\pi3=3\sqrt3\)
      • \(y-3\sqrt3=3\left(x-\frac\pi3\right)\), so \(y=3x-\pi+3\sqrt3\)

    2. (b) \(y=e^x-5x\): the tangent perpendicular to \(x+3y=6\).

      • Line \(x+3y=6\) has gradient \(-\frac13\), so the tangent has gradient 3
      • \(y'=e^x-5=3\), so \(x=\ln8\) and \(y=8-5\ln8\)
      • \(y-(8-5\ln8)=3(x-\ln8)\), so \(y=3x+8-8\ln8=3x+8-24\ln2\)

  6. Find the interval in which \(f(x)=2\ln x-10x\) is a decreasing function.

    • \(f'(x)=\frac2x-10\)
    • Decreasing when \(\frac2x-10<0\), so \(\frac2x<10\)
    • \(x>0\) for \(\ln x\), so \(x>\frac15\)

In context

  1. A weather balloon descends after its gas is released. Its height above the ground is \(h=\frac{6t+54}{\sqrt t}\) metres, \(t\) minutes after the release.

    1. (a) Find the rate of change of height when \(t=4\).

      • \(h=6t^{1/2}+54t^{-1/2}\)
      • \(h'=3t^{-1/2}-27t^{-3/2}\)
      • At \(t=4\): \(\frac32-\frac{27}{8}=1.5-3.375=-1.875\) m/min

    2. (b) What does the sign of your answer tell you?

      • The rate is negative, so the height is decreasing.

  2. The number of bacteria in a culture, in millions, is modelled by \(N=(3+\sqrt t)^2\), where \(t\) is the time in hours after the culture is set up.

    1. (a) What is the number of bacteria at the start?

      • \(N(0)=(3+0)^2=9\)

    2. (b) How long does it take to reach 36 million?

      • \((3+\sqrt t)^2=36\), so \(3+\sqrt t=6\)
      • \(\sqrt t=3\), so \(t=9\)

    3. (c) Find the rate of increase when \(t=16\).

      • \(N=9+6t^{1/2}+t\)
      • \(N'=3t^{-1/2}+1\)
      • At \(t=16\): \(\frac34+1=1.75\) million per hour

    4. (d) State one limitation of the model.

      • Any one limitation of the model, such as growth continuing forever.

Stretch

  1. Find the coordinates of the point of tangency.

    1. (a) The line \(y=\dfrac{x}{12}+3\) is a tangent to \(y=\sqrt x\).

      • \(y'=\frac1{2\sqrt x}=\frac1{12}\), so \(\sqrt x=6\) and \(x=36\)
      • \(y=\sqrt{36}=6\)
      • Check: \(\frac{36}{12}+3=6\)

    2. (b) The line \(3y+3\sqrt3x=3+\sqrt3\pi\) is a tangent to \(y=2\cos x\), \(0<x<\dfrac\pi2\), at P.

      • Gradient of \(3y+3\sqrt3x=3+\sqrt3\pi\) is \(-\sqrt3\)
      • \(y'=-2\sin x=-\sqrt3\), so \(\sin x=\frac{\sqrt3}2\) and \(x=\frac\pi3\) (in the interval)
      • \(y=2\cos\frac\pi3=1\)

  2. The curve is \(y=5x-e^x\).

    1. (a) Find the equation of the normal at the point where \(x=\ln\dfrac{26}{5}\).

      • \(y'=5-e^x=5-\frac{26}{5}=-\frac15\) at \(x=\ln\frac{26}5\)
      • Normal gradient 5; \(y=5\ln\frac{26}{5}-\frac{26}{5}\)
      • \(y-y_1=5(x-\ln\frac{26}5)\), so \(y=5x-\frac{26}{5}\)

    2. (b) Show that this normal does not meet the curve at any other point.

      • \(5x-e^x=5x-\frac{26}{5}\), so \(e^x=\frac{26}{5}\)
      • \(e^x\) takes each value once, so \(x=\ln\frac{26}5\) only
      • The normal does not meet the curve at any other point

  3. The curve is \(y=3\tan x\). Find the equation of the tangent

    1. (a) at the point where \(x=k\);

      • \(y'=\frac3{\cos^2x}\)
      • At \(x=k\): \(m=\frac3{\cos^2k}\), \(y_1=3\tan k\)
      • \(y-3\tan k=\frac{3}{\cos^2k}(x-k)\)

    2. (b) when \(k=\dfrac\pi6\).

      • \(\cos^2\frac\pi6=\frac34\), so \(m=4\); \(y_1=3\tan\frac\pi6=\sqrt3\)
      • \(y-\sqrt3=4\left(x-\frac\pi6\right)\), so \(y=4x+\sqrt3-\frac{2\pi}3\)

  4. The tangent to \(y=\sqrt x\) at \(x=k\) is perpendicular to the tangent to \(y=\dfrac{36}{\sqrt x}\), also at \(x=k\). Find k.

    • \(\frac{d}{dx}\sqrt x=\frac1{2\sqrt k}\) and \(\frac{d}{dx}36x^{-1/2}=-18k^{-3/2}\)
    • Perpendicular: \(\frac1{2\sqrt k}\times(-18k^{-3/2})=-1\), so \(\frac9{k^2}=1\)
    • \(k^2=9\), so \(k=3\) (k > 0)

  5. The curve \(y=5\cos x\), for \(-\pi<x<\pi\), meets the x-axis at P and Q. The tangents at P and Q meet at R.

    −3−2−1123−22468xyy = 5cos xPQR

    1. (a) Find the area of triangle PQR.

      • \(P\left(-\frac\pi2,0\right)\), \(Q\left(\frac\pi2,0\right)\); \(y'=-5\sin x\) gives gradients 5 and −5
      • Tangents meet on the y-axis at \(R\left(0,\frac{5\pi}2\right)\)
      • Area \(=\frac12\times\pi\times\frac{5\pi}2=\frac{5\pi^2}{4}=12.3\)

    2. (b) The curve \(y=4\sin x\), for \(0\le x<2\pi\), meets the x-axis at A and B. The tangents at A and B meet at C. Find the area of triangle ABC.

      • \(y'=4\cos x\): at A(0, 0) gradient 4, at B(\(\pi\), 0) gradient −4
      • Tangents \(y=4x\) and \(y=-4(x-\pi)\) meet at \(C\left(\frac\pi2,2\pi\right)\)
      • Area \(=\frac12\times\pi\times2\pi=\pi^2=9.87\)

  6. A tangent to the curve \(y=\sqrt x\) passes through the point \((-5,2)\).

    1. (a) Find the coordinates of the point where the tangent touches the curve.

      • Tangent at \(x=a\): \(y=\sqrt a+\frac1{2\sqrt a}(x-a)\)
      • Through \((-5,2)\): \(2=\sqrt a+\frac{-5-a}{2\sqrt a}\), so \(4\sqrt a=a-5\), which is \(a-4\sqrt a-5=0\)
      • \(\sqrt a=5\) (the root \(-1\) is not allowed), so \(a=25\) and the point is \((25,5)\)

    2. (b) Find the equation of the tangent.

      • Gradient \(\frac1{2\times5}=\frac1{10}\)
      • \(y-5=\frac1{10}(x-25)\), so \(y=\frac x{10}+\frac52\)

jmaths.xyz · AI HL · AHL 2.2 Tangents, normals and gradient problems