BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 6P
To determine

To find: The points on the parabola y=1x2 at which the tangent line cuts from first quadrant the triangle with the smallest area.

Expert Solution

Answer to Problem 6P

The point on the parabola is (13,23)

Explanation of Solution

Let the points on the curve be (u,v) .

As the points lie on the parabola y=1x2 write v=1u2 .

Slope of the tangent line is, dydx=2x .

Slope of the tangent line at point (u,v) it is 2u .

Write the equation of the tangent line by using slope-point form.

Thus, the equation of the tangent line is, yv=2u(xu) .

Find x-intercepts and y-intercepts of the tangent.

To find x-intercept, put y=0 .

Thus, the x-intercept is u+v2u .

Similarly to find y-intercept, put x=0 .

Thus, the y-intercept is v+2u2 .

Now area of the triangle formed is given by,

A=12xinterceptyintercept=12(u+v2u)(v+2u2)=(2u2+v)24u

Substitute value of v as v=1u2 .

A=(2u2+v)24u=(2u2+1u2)24u=(1+u2)24u

Differentiate A with respect to u.

dAdu=ddu((1+u2)24u)=16u2(1+u2)(1+u2)2416u2

Set dAdu=0 to get,

16u2(1+u2)(1+u2)2416u2=016u2(1+u2)(1+u2)24=016u2(1+u2)=(1+u2)2416u2=(1+u2)4

Simplify further as follows.

16u2=4(1+u2)12u2=4u2=412u=±13

As the point is in first quadrant u=13 .

Substitute to get value of v.

v=1u2=1(13)2=113=23

Hence, for the values u=13 and v=23 the tangent line to the curve y=1x2 cuts the first quadrant in a triangle with smallest area.

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