Finding Tangents to Curves Using Derivatives

Date: 09/29/1999 at 22:58:04
From: Emon Eshaque
Subject: Calculus

1. Let T be the point on the graph y = k/x whose x-coordinate is t. 
   Prove that the x-intercept of the tangent at T is 2t.

2. The line y = mx+1 meets the parabola y = (1/4)x^2 at A and B. Prove 
   that the tangents at A and B are perpendicular.

3. A curve known as a strophoid is defined implicitly by the equation

     x3 + 4x2 + xy2 - 4y2 = 0   (where -4 < x < 4)

   and is shown in the diagram. Find the points at which the tangent 
   line is horizontal. Round the coordinates off to 2 decimal places. 

I will be glad if you send me some hints.

Date: 10/24/1999 at 22:16:41
From: Doctor Jaffee
Subject: Re: Calculus

Hi Emon,

Here are some hints that should help you.

In problem 1 we are concerned about the point where the x-coordinate 
is t. If you substitute t for x in the equation, you will have the 
y-coordinate at that point. The derivative of the function will tell 
you the slope of the tangent line, so calculate the derivative and 
then substitute t for x to find the slope at your point. Now you know 
the slope and a point on your line, so use the slope-point form of the 
equation to determine the particular equation. Finally, the 
x-intercept is the point where y = 0. Therefore, substitute 0 for y 
and solve for x and you should get 2t.

In problem 2, remember that two lines are perpendicular if the product 
of their slopes is -1.

Problem 3:

      d/dx(x^3) = 3x^2
     d/dx(4x^2) = 8x
     d/dx(xy^2) = y^2 + 2xyy'
        d(4y^2) = 8yy'

Take the derivative of both sides of the equation and solve for y'. 
Since the slope of a horizontal line is 0, set y' = 0.

I hope these hints are helpful. Give them a try, and if you need more 
help, write back.

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/