## Exercises2.4Tangent Lines and Implicit Differentiation

Solve the following problems.

###### 1.
1. Find $\ds \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$ where $\ds f(x)=\frac{3x+1}{x-2}\text{.}$

2. What does the result in (a) tell you about the tangent line to the graph of $y=f(x)$ at $x=1\text{?}$

3. Find the equation of the tangent line to $y=f(x)$ at $x=1\text{.}$

1. $f^\prime(1)=-7\text{.}$

2. The slope is equal to $-7\text{.}$

3. $y=-7x+3\text{.}$

###### 2.

Consider a tangent of the curve $y=17-2x^2$ that goes through the point $(3,1)\text{.}$

1. Provide a diagram of this situation. Can you draw two tangent lines?

2. Find the slopes of those tangent lines.

3. Find the equation of those tangent lines.

1. 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 $k=-8\text{.}$

2. $y=-16x+49$ and $y=-8x+25\text{.}$

###### 3.

The Figure below shows a circle with the radius 1 inscribed in the parabola $y=x^2\text{.}$ Find the centre of the circle.

Hint
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.
$\ds a=\frac{5}{4}\text{.}$

Given that $\ds \sinh x=\frac{e^x-e^{-x}}{2}$ and $\ds \cosh x =\frac{e^x+e^{-x}}{2}\text{:}$

###### 4.

Prove that $\cosh(x+y)=\cosh x \cosh y +\sinh x\sinh y\text{.}$

###### 5.

Find $\ds \lim_{x\to \infty}\tanh x\text{.}$

$1\text{.}$
###### 6.

Find the equation of the tangent line to the curve $y=4+3x+\cosh x$ at the point $(0,5)\text{.}$

$y=3x+5\text{.}$
###### 7.

At what point on the curve $y=\sinh x$ does the tangent line have a slope of 1?

Hint
Solve $\ds y'=\cosh x=1\text{.}$
The point is $(0,0)\text{.}$
###### 8.

Find the point(s) on the curve $y=x^3$ where the line through the point $(4,0)$ is tangent to the curve.

Hint
Solutions of $-a^3=3a^2(4-a)$ are $a=0$ and $a=6\text{.}$
The points are $(0,0)$ and $(6,216)\text{.}$
###### 9.

Find the equation of the tangent line to $y=x^2-1$ that has slope equal to $-4\text{.}$

$y=-4x-5\text{.}$
###### 10.

Find the equation of the tangent line to $y=x^2$ passing through the point $(0,-3)\text{.}$

$y=2\sqrt{3}x\pm 3\text{.}$
###### 11.
1. Find $\ds \arcsin \left( -\frac{1}{\sqrt{2}}\right)$

2. Find the equation of the tangent line to the graph of $y=\arcsin x$ when $\ds x=-\frac{1}{\sqrt{2}}\text{.}$

1. $\ds -\frac{\pi }{4}\text{.}$

2. $\ds y =\sqrt{2}x +1-\frac{\pi }{4}\text{.}$

###### 12.

Find the equation of the line that is tangent to the graph of $\ds y=\sqrt{x}-\frac{1}{\sqrt{x}}$ at $x=1\text{.}$

$y=x-1\text{.}$
###### 13.

Find the equation of the line that is tangent to the graph of $\ds y=e^{x^2}$ at $x=2\text{.}$

$y=e^4(4x-7)\text{.}$
###### 14.

Find the values of $c$ such that the line $\ds y=\frac{3x}{2}+6$ is tangent to the curve $y=c\sqrt{x}\text{.}$

$c=\pm6\text{.}$
###### 15.

Find the values of $a$ such that the tangent line to $\ds f(x)=\frac{2x^2}{1+x^2}$ at the point $(a,f(a))$ is parallel to the tangent of $g(x)=\tan^{-1}(x)$ at $(a,g(a))\text{.}$

$a= 2\pm\sqrt{3}\text{.}$
###### 16.

Consider the function $\ds y=3x^2+2x-10$ on the interval $[1,5]\text{.}$ Does this function have a tangent line with a slope of 20 anywhere on this interval? Explain.

Hint

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

Yes.
###### 17.

Let $C$ be the curve $y=(x-1)^3$ and let $L$ be the line $3y+x=0\text{.}$

1. Find the equation of all lines that are tangent to $C$ and are also perpendicular to $L\text{.}$

2. Draw a labeled diagram showing the curve $C\text{,}$ the line $L\text{,}$ and the line(s) of your solution to part (a). For each line of your solution, mark on the diagram the point where it is tangent to $C$ and (without necessarily calculating the coordinates) the point where it is perpendicular to $L\text{.}$

1. The lines are $y=3x-1$ and $y=3x-5\text{.}$

Solution
1. 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 $y=3x-5\text{.}$

###### 18.

Find the value of $h^\prime(0)$ if $h(x)+x\cos(h(x))=x^2+3x\text{.}$

$h^\prime(0)= 2\text{.}$
###### 19.

Find the derivative of $y$ with respect to $x$ for the curve $e^y\ln (x+y)+1=\cos (xy)$ at the point $(1,0)\text{.}$

$-1\text{.}$
Solution

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 $\ds \left. \frac{dy}{dx}\right| _{x=1}=-1\text{.}$

Find $\ds \frac{dy}{dx}$ if:

###### 20.

$x^5+y^5=5xy$

$\ds \frac{dy}{dx}=\frac{y-x^4}{y^4-x}\text{.}$
###### 21.

$x^2+x^3y-xy^2=1$

$\ds y^\prime=\frac{-3x^2 y-2x+y^2}{x^3-2xy}\text{.}$
###### 22.

$x^y=y^x$

$\ds \frac{dy}{dx}=\frac{y(x\ln y-y)}{x(y\ln x -x)}.$
Solution
\begin{equation*} \begin{split} \ds y\ln x \amp=x\ln y\\ \frac{dy}{dx}\cdot \ln x+\frac{y}{x}\amp=\ln y+\frac{x}{y}\cdot \frac{dy}{dx}\\ \frac{dy}{dx}\amp=\frac{y(x\ln y-y)}{x(y\ln x -x)}. \end{split} \end{equation*}
###### 23.

$e^y-3^x=x\sinh y$

$\ds \frac{dy}{dx}=\frac{3^x\ln 3 +\sinh y}{e^y-x\cosh y}\text{.}$
###### 24.

$y-2ye^{xy}=0$

$\ds y^\prime=\frac{2y^2 e^xy}{1-2xye^xy-2e^xy}\text{.}$
###### 25.

$\sinh x-\cos y=x^2y$

$\ds \frac{dy}{dx}=\frac{\cosh x -2xy}{x^2-\sin y}\text{.}$
###### 26.

$\ln (x-y)=xy+y^3$

$\ds \frac{dy}{dx}=\frac{1-y(x-y)}{1+(x-y)(x+3y^2)}\text{.}$
###### 27.

$\ds \ln y +x=x^2+x\cos y$

$\ds y^\prime = \frac{y(2x+\cos (y)-1}{xy\sin(y)+1}\text{.}$
###### 28.

$\ds x\sin y+y\cos x=1$

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

$\ds \sin y+x=x^2+x\cos y$

$\ds y^\prime = \frac{2x+\cos (y)-1}{x\sin(y)+\cos (y)}\text{.}$
###### 30.

$\ds \cos x+e^y=x^2+\tan^{-1}y$

$\ds y^\prime = \frac{(y^2+1)(2x+\sin? (x))}{e^y y^2+e^y-1}\text{.}$
###### 31.

$\ds y\cos (x^2)=x\sin(y^2)$

$\ds y^\prime = \frac{2xy\sin(x^2 )+\sin(y^2)}{\cos (x^2 )-2xy\cos(y^2)}\text{.}$
###### 32.

$\ds \tan y+\ln x=x^2+\tan^{-1}y$

$\ds y^\prime = \frac{(2x^2-1)(y^2+1)}{x(y^2 \sec^2 (y)+\tan^2 (y))}\text{.}$
###### 33.

$\ds (xy+1)^2=\tan (x+y^3)$

$\ds y^\prime = \frac{\sec^2 (x+y^3 )-2y(2xy^ +1)}{3y^2 \sec^2 (x+y^3 )-2x(xy+1)}\text{.}$
###### 34.

Find the slope of the tangent line to the curve $y+x\ln y-2x=0$ at the point $(1/2,1)\text{.}$

$\ds \dfrac{4}{3}\text{.}$

Use implicit differentiation to answer the following:

###### 35.

Find the tangent line to the graph of $\sin (x+y)=y^2\cos x$ at $(0,0)\text{.}$

$x+y=0\text{.}$
###### 36.

Show that the tangent lines to the graph of $x^2-xy+y^2=3\text{,}$ at the points where the graph crosses the $x$-axis, are parallel to each other.

Solution
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 $y'=2\text{.}$
###### 37.

The curve implicitly defined by

\begin{equation*} x\sin y+y\sin x=\pi \end{equation*}

passes through the point $\ds P=\left( \frac{\pi }{2},\frac{\pi }{2}\right)\text{.}$

1. Find the slope of the tangent line through $P$

2. Write the tangent line through $P\text{.}$

1. $\displaystyle -1$

2. $\displaystyle y= \pi-x$

###### 38.

Write the equation of the line tangent to the curve $\ds \sin (x+y)=xe^{x+y}$ at the origin $(0,0)\text{.}$

$y= 0 \text{.}$
###### 39.

Write the equation of the line tangent to the curve $\ds \sin (x+y)=2x-2y$ at the point $(\pi,\pi)\text{.}$

$\ds y-\pi=\frac{1}{3}(x-\pi)\text{.}$
###### 40.

Find the slope of the tangent line to the curve $\ds xy =6e^{2x-3y}$ at the point $(3,2)\text{.}$

$\dfrac{10}{21}.$
###### 41.

Find the $x$-coordinates of points on the curve $\ds x^3+y^3=6xy$ where the tangent line is vertical.

$x=0\text{,}$ $x=2\sqrt{3}{4}\text{.}$
###### 42.
1. Find $\ds \frac{dy}{dx}$ for the function defined implicitly by $x^2y+ay^2=b\text{,}$ where $a$ and $b$ are fixed constants.

2. For the function defined in part (a) find the values of the constants $a$ and $b$ if the point $(1,1)$ is on the graph and the tangent line at $(1,1)$ is $4x+3y=7\text{.}$

1. $\ds-\frac{2xy}{x^2+2ay}\text{.}$

2. $\ds a=\frac{1}{4}$ and $\ds b=\frac{5}{4}\text{.}$

Solution
1. Let $a \text{,}$ $b$ be constants. Then we have

\begin{equation*} \begin{split} \frac{d}{dx} \left(x^2 y + ay^2 \right) \amp = \frac{d}{dx} \left( b \right) \\ \frac{d}{dx} \left(x^2 y\right) + \frac{d}{dx} \left(ay^2\right) \amp = 0\\ 2xy + x^2\frac{dy}{dx} + 2ay \frac{dy}{dx} \amp = 0\\ \frac{dy}{dx} \left(x^2+2ay\right) \amp = -2xy\\ \frac{dy}{dx} \amp = \frac{-2xy}{x^2+2ay}. \end{split} \end{equation*}
2. We solve the system of equations:

\begin{equation*} \begin{split}1+a=\amp b \\ -\frac{2}{1+2a}= \amp -\frac{4}{3} \end{split} \end{equation*}

to get $\ds a=\frac{1}{4}$ and $\ds b=\frac{5}{4}\text{.}$

###### 43.

Let $l$ be any tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{k}\text{,}$ $k>0\text{.}$ Show that the sum of the $x$-intercept and the $y$-intercept of $l$ is $k\text{.}$

$(a+\sqrt{ab})+(b+\sqrt{ab})=(\sqrt{a}+\sqrt{b})^2=k\text{.}$
Solution

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 $(a+\sqrt{ab})+(b+\sqrt{ab})=(\sqrt{a}+\sqrt{b})^2=k\text{.}$

###### 44.

Show that the length of the portion of any tangent line to the curve

\begin{equation*} x^{2/3}+y^{2/3}=9\text{,} \end{equation*}

cut off by the coordinate axis is constant. What is this length?

$\sqrt{9^3}=27\text{.}$
Solution
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
\begin{equation*} \begin{split} \left( a+\sqrt[3]{ab^2}\right) ^2+\left( b+\sqrt[3]{a^2b}\right) ^2 \amp =a^2+2a\sqrt[3]{ab^2}+ b\sqrt[3]{a^2b} + b^2+2b\sqrt[3]{a^2b}+a\sqrt[3]{ab^2} \\ \amp=\left( \sqrt[3]{a^2}+\sqrt[3]{b^2}\right) ^3\\ \amp=9^3 \end{split}. \end{equation*}
The length of the portion is $\sqrt{9^3}=27\text{.}$
###### 45.

Let $C$ denote the circle whose equation is $(x-5)^2+y^2=25\text{.}$ Notice that the point $(8,-4)$ lies on the circle $C\text{.}$ Find the equation of the line that is tangent to $C$ at the point $(8,-4)\text{.}$

$\ds y+4=\frac{3}{4}(x-8)\text{.}$
###### 46.

The so called devil's curve is described by the equation

\begin{equation*} y^2(y^2-4)=x^2(x^2-5)\text{.} \end{equation*}
1. Compute the $y$-intercepts of the curve.

2. Use implicit differentiation to find an expression for $\ds \frac{dy}{dx}$ at the point $(x,y)\text{.}$

3. Give an equation for the tangent line to curve at $(\sqrt{5},0)\text{.}$

1. $(0,0)\text{,}$ $(0,\pm 2)\text{.}$

2. $\ds y^\prime =\frac{x(2x^2-5)}{2y(y^2-2)}\text{.}$

3. $x=\sqrt{5}\text{.}$

###### 47.

The equation $e^y+y(x-2)=x^2-8$ defines $y$ implicitly as a function of $x$ near the point $(3,0)\text{.}$

1. Determine the value of $y'$ at this point.

2. Use the linear approximation to estimate the value of $y$ when $x=2.98\text{.}$

1. $\displaystyle y'(3) = 3.$

2. $\displaystyle y(2.98)\approx 3\cdot 2.98-9=-0.06$

###### 48.

The equation $e^y+y(x-3)=x^2-15$ defines $y$ implicitly as a function of $x$ near the point $A(4,0)\text{.}$

1. Determine the values of $y'$ and $y''$ at this point.

2. Use the tangent line approximation to estimate the value of $y$ when $x=3.95\text{.}$

3. Is the true value of $y$ greater or less than the approximation in part (b)? Make a sketch showing how the curve relates to the tangent line near the point $A(4,0)\text{.}$

1. $y'(4)=4$ and $y"(4)=-11$
2. $y(3.95)\approx -0.2\text{.}$