Tangents, normals and gradient problems Problem set A Worked solutions

Differentiate with respect to x.

1. (a) \(y=7e^x\)

   • The derivative of \(e^x\) is \(e^x\)
   • Multiplying constant stays: \(\frac{dy}{dx}=7e^x\)

2. (b) \(y=-4e^x\)

   • \(\frac{dy}{dx}=-4e^x\)

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

   • \(y=-\frac{e^x}{3}\) is \(-\frac13e^x\)
   • \(\frac{dy}{dx}=-\frac13e^x\)

4. (d) \(y=\dfrac{5e^x}{2}\)

   • \(\frac{dy}{dx}=\frac52e^x\)

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

1. (a) \(y=\sin x\) at \(x=\dfrac{\pi}{6}\).

   • \(y'=\cos x\)
   • At \(x=\frac\pi6\): \(\cos\frac\pi6=\frac{\sqrt3}{2}\)

2. (b) \(y=3e^x-2x+1\) at \(x=\ln 4\).

   • \(y'=3e^x-2\)
   • At \(x=\ln4\): \(3\times4-2=10\)

Find the point on the curve \(y=3+4x-e^x\) where the gradient is 1. Give exact coordinates, then 3 significant figures.

• \(y'=4-e^x=1\), so \(e^x=3\) and \(x=\ln3\)
• \(y=3+4\ln3-3=4\ln3\)
• Point \((\ln3,\,4\ln3)=(1.10,\,4.39)\)

Find the equation of the tangent to the curve \(y=x^2-6x+5\) that is parallel to the line \(y=4x-3\).

• \(y'=2x-6=4\), so \(x=5\)
• \(y=25-30+5=0\): the point is \((5,0)\)
• \(y-0=4(x-5)\), so \(y=4x-20\)

Find the equation of each tangent.

1. (a) \(y=4\cos x\), \(0<x<\dfrac\pi2\): the tangent parallel to \(2y+4x=9\).

   • Parallel to \(2y+4x=9\) (gradient −2): \(y'=-4\sin x=-2\), so \(\sin x=\frac12\) and \(x=\frac\pi6\)
   • \(y=4\cos\frac\pi6=2\sqrt3\)
   • \(y-2\sqrt3=-2\left(x-\frac\pi6\right)\), so \(y=-2x+\frac\pi3+2\sqrt3\)

2. (b) \(y=e^x-3x\): the tangent perpendicular to \(x+2y=8\).

   • Line \(x+2y=8\) has gradient \(-\frac12\), so the tangent has gradient 2
   • \(y'=e^x-3=2\), so \(x=\ln5\) and \(y=5-3\ln5\)
   • \(y-(5-3\ln5)=2(x-\ln5)\), so \(y=2x+5-5\ln5\)

Find the interval in which \(f(x)=\ln x-4x\) is a decreasing function.

• \(f'(x)=\frac1x-4\)
• Decreasing when \(\frac1x-4<0\), so \(\frac1x<4\)
• \(x>0\) for \(\ln x\), so \(x>\frac14\)

A drone descends to land. Its height above the ground is \(h=\frac{4t+36}{\sqrt t}\) metres, \(t\) seconds after it starts to descend.

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

   • \(h=4t^{1/2}+36t^{-1/2}\)
   • \(h'=2t^{-1/2}-18t^{-3/2}\)
   • At \(t=4\): \(\frac22-\frac{18}{8}=1-2.25=-1.25\) m/s

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

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

The number of algae cells in a pond, in millions, is modelled by \(N=(2+\sqrt t)^2\), where \(t\) is the time in days after the pond is first measured.

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

   • \(N(0)=(2+0)^2=4\)

2. (b) How long does it take to reach 25 million?

   • \((2+\sqrt t)^2=25\), so \(2+\sqrt t=5\)
   • \(\sqrt t=3\), so \(t=9\)

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

   • \(N=4+4t^{1/2}+t\)
   • \(N'=2t^{-1/2}+1\)
   • At \(t=16\): \(\frac24+1=1.5\) million per day

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

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

Find the coordinates of the point of tangency.

1. (a) The line \(y=\dfrac{x}{10}+\dfrac52\) is a tangent to \(y=\sqrt x\).

   • \(y'=\frac1{2\sqrt x}=\frac1{10}\), so \(\sqrt x=5\) and \(x=25\)
   • \(y=\sqrt{25}=5\)
   • Check: \(\frac{25}{10}+\frac52=5\)

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

   • Gradient of \(3y=3x+3\sqrt3-\pi\) is 1
   • \(y'=2\cos x=1\), so \(\cos x=\frac12\) and \(x=\frac\pi3\) (in the interval)
   • \(y=2\sin\frac\pi3=\sqrt3\)

The curve is \(y=3x-e^x\).

1. (a) Find the equation of the normal at the point where \(x=\ln\dfrac{10}{3}\).

   • \(y'=3-e^x=3-\frac{10}{3}=-\frac13\) at \(x=\ln\frac{10}3\)
   • Normal gradient 3; \(y=3\ln\frac{10}{3}-\frac{10}{3}\)
   • \(y-y_1=3(x-\ln\frac{10}3)\), so \(y=3x-\frac{10}{3}\)

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

   • \(3x-e^x=3x-\frac{10}{3}\), so \(e^x=\frac{10}{3}\)
   • \(e^x\) takes each value once, so \(x=\ln\frac{10}3\) only
   • The normal does not meet the curve at any other point

The curve is \(y=2\tan x\). Find the equation of the tangent

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

   • \(y'=\frac2{\cos^2x}\)
   • At \(x=k\): \(m=\frac2{\cos^2k}\), \(y_1=2\tan k\)
   • \(y-2\tan k=\frac{2}{\cos^2k}(x-k)\)

2. (b) when \(k=\dfrac\pi4\).

   • \(\cos^2\frac\pi4=\frac12\), so \(m=4\); \(y_1=2\)
   • \(y-2=4\left(x-\frac\pi4\right)\), so \(y=4x+2-\pi\)

The tangent to \(y=\sqrt x\) at \(x=k\) is perpendicular to the tangent to \(y=\dfrac{16}{\sqrt x}\), also at \(x=k\). Find k.

• \(\frac{d}{dx}\sqrt x=\frac1{2\sqrt k}\) and \(\frac{d}{dx}16x^{-1/2}=-8k^{-3/2}\)
• Perpendicular: \(\frac1{2\sqrt k}\times(-8k^{-3/2})=-1\), so \(\frac4{k^2}=1\)
• \(k^2=4\), so \(k=2\) (k > 0)

The curve \(y=3\cos x\), for \(-\pi<x<\pi\), meets the x-axis at P and Q. The tangents at P and Q meet at R.

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

   • \(P\left(-\frac\pi2,0\right)\), \(Q\left(\frac\pi2,0\right)\); \(y'=-3\sin x\) gives gradients 3 and −3
   • Tangents meet on the y-axis at \(R\left(0,\frac{3\pi}2\right)\)
   • Area \(=\frac12\times\pi\times\frac{3\pi}2=\frac{3\pi^2}{4}=7.40\)

2. (b) The curve \(y=2\sin x\), 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.

   • \(y'=2\cos x\): at A(0, 0) gradient 2, at B(\(\pi\), 0) gradient −2
   • Tangents \(y=2x\) and \(y=-2(x-\pi)\) meet at \(C\left(\frac\pi2,\pi\right)\)
   • Area \(=\frac12\times\pi\times\pi=\frac{\pi^2}{2}=4.93\)

A tangent to the curve \(y=\sqrt x\) passes through the point \((-8,1)\).

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 \((-8,1)\): \(1=\sqrt a+\frac{-8-a}{2\sqrt a}\), so \(2\sqrt a=a-8\), which is \(a-2\sqrt a-8=0\)
   • \(\sqrt a=4\) (the root \(-2\) is not allowed), so \(a=16\) and the point is \((16,4)\)

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

   • Gradient \(\frac1{2\times4}=\frac18\)
   • \(y-4=\frac18(x-16)\), so \(y=\frac x8+2\)