Welcome to our community

Be a part of something great, join today!

Paul's questions at Yahoo! Answers regarding tangent and normal lines

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here are the questions:

Equation of tangent and normal?

1.find the equation of the tangent and normal to the curve y = x/a+a/x at (a/2, 5/2)

2. at what point of the parabola y=x^2-3x-5 is the tangent line parallel to 3x-y=2? find the eqn of the tangent line.

3. find the eqn. of the line normal to the curve y=3x^5+10x^3+15x+1 at its point of inflection.

4. find the critical points and the points of inflection of the curve y= 3x^4-8x^3+6x^2


please show your solution. help me. thanks!
Here is a link to the questions:

Equation of tangent and normal? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Paul,

1.) We are asked to find the equation of the tangent and normal lines to the curve \(\displaystyle y=\frac{x}{a}+\frac{a}{x}\) at \(\displaystyle \left(\frac{a}{2},\frac{5}{2} \right)\).

First, we need to compute the derivative of the curve to find the slopes:

\(\displaystyle \frac{dy}{dx}=\frac{1}{a}-\frac{a}{x^2}=\frac{x^2-a^2}{ax^2}\)

Hence, the slope of the tangent line is:

\(\displaystyle m=\left.\frac{dy}{dx} \right|_{x=\frac{a}{2}}=\frac{\left(\frac{a}{2} \right)^2-a^2}{a\left(\frac{a}{2} \right)^2}=-\frac{3}{a}\)

and so the slope of the normal line is:

\(\displaystyle -\frac{1}{m}=\frac{a}{3}\)

We now have the slopes and the point, so applying the point-slope formula, we find:

i) Tangent line:

\(\displaystyle y-\frac{5}{2}=-\frac{3}{a}\left(x-\frac{a}{2} \right)\)

\(\displaystyle y=-\frac{3}{a}x+4\)

ii) Normal line:

\(\displaystyle y-\frac{5}{2}=\frac{a}{3}\left(x-\frac{a}{2} \right)\)

\(\displaystyle y=\frac{a}{3}x+\frac{15-a^2}{6}\)

Here are some plots with a few values for $a$:

paul1.jpg

2.) We are asked to find at what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find the equation of the tangent line.

First we observe that we may arrange the given line in slope-intercept form as follows:

\(\displaystyle y=3x-2\)

Now, we want to equate the derivative of the parabola to the slope of the line (since parallel lines have equal slopes):

\(\displaystyle 2x-3=3\)

\(\displaystyle x=3\)

Now, to find the $y$-coordinate:

\(\displaystyle y(3)=(3)^2-3(3)-5=-5\)

and so we now have the slope \(\displaystyle m=3\) and the point \(\displaystyle (3,-5)\), and applying the point-slope formula, we find then that the equation of the tangent line parallel to the given line is:

\(\displaystyle y-(-5)=3(x-3)\)

\(\displaystyle y=3x-14\)

Here is a plot of the parabola, the given line, and the line tangent to the parabola and parallel to the given line:

paul2.jpg

3.) We are asked to find the equation of the line normal to the curve $y=3x^5+10x^3+15x+1$ at its point of inflection.

From the first derivative, we may determine the slope of the tangent line, once we know where the point of inflection is.

\(\displaystyle \frac{dy}{dx}=15x^4+30x^2+15=15\left(x^2+1 \right)^2\)

\(\displaystyle \frac{d^2y}{dx^2}=60x^3+60x=60x\left(x^2+1 \right)=0\)

We find that the second derivative has only the real root $x=0$ and since it is of multiplicity 1, we know the second derivative changes sign across this root, hence the point of inflection is at \(\displaystyle (0,1)\). Now the slope of the normal line is:

\(\displaystyle m=-\frac{1}{15}\) and so, applying the point slope formula, we find the normal line is:

\(\displaystyle y-1=-\frac{1}{15}(x-0)\)

\(\displaystyle y=-\frac{1}{15}x+1\)

4.) We are asked to find the critical points and the points of inflection of the curve $y=3x^4-8x^3+6x^2$.

Critical or stationary points:

We want to equate the first derivative to zero, and solve for $x$:

\(\displaystyle \frac{dy}{dx}=12x^3-24x^2+12x=12x(x-1)^2=0\)

The critical values are then:

\(\displaystyle x=0,\,1\)

Because the root $x=1$ is of multiplcity 2, we may conclude that this is not an extremum, and by the first derivative test we know the critical value $x=0$ is at a relative minimum. Thus \(\displaystyle (0,0)\) is a relative minimum.

Points of inflection:

\(\displaystyle \frac{d^2y}{dx^2}=36x^2-48x+12=12\left(3x^2-4x+1 \right)=12(3x-1)(x-1)=0\)

Since the roots are of multiplcity 1, we know the sign of the second derivative changes across then, and so we may conclude that the points of inflection are at:

\(\displaystyle \left(\frac{1}{3},\frac{11}{27} \right)\)

\(\displaystyle (1,1)\)

Here is a plot of the quartic function, showing its minimum at the origin and the points of inflection:

paul4.jpg

To Paul and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.