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 15P

(a)

To determine

To show: If P(a,a2) is any point on the parabola y=x2 but not at the origin and Q is any point on the parabola where the normal line at P intersects the parabola, then the y- coordinate of Q is smallest when a=12

Expert Solution

Explanation of Solution

The equation of the parabola is given as y=x2.

On differentiating with respect to x, dydx=2x.

Thus, the slope of the tangent is, dydx=2x.

So, the slope of the normal is, 12x.

At the point (a,a2), the slope of tangent is 2a and the slope of normal is 12a.

By slope point form, the equation of normal is ya2=12a(xa).

Use y=x2 in the equation of normal.

ya2=12a(xa)x2a2=12a(xa)(x+14a)2=(a+14a)2x=a or (a12a)

One is the x-coordinate of P say xP=a and other is x-coordinate of Q say xQ=(a+12a).

Using the value of xQ in the equation of parabola, write yQ=(a+12a)2.

Find the derivative of yQ with respect to a as follows.

dyQda=2(a+12a)(1+12a2)

Set dyQda=0,

dyQda=02(a+12a)(1+12a2)=0(2a+1a)(2+1a2)=0(2a2+1a)(2a21a2)=02a2+1=0 or 2a21=0

On simplification,

2a2+1=0 or 2a21=0a=12 or a=12a=12 or a=12

Since a is a real number, the value of a is a=12.

Compute the second derivative of yQ with respect to a as follows.

d2yQda2=dda(2(a+12a)(1+12a2))=2{(a+12a)dda(1+12a2)+(1+12a2)dda(a+12a)}[By product rule]=2{(a+12a)(1a3)+(1+12a2)(112a2)}=4a4+32a4

Clearly for any value of a, d2yQda2>0.

That is, the value of y coordinate is minimum when a=12.

Hence the proof.

(b)

To determine

To show: If P(a,a2) is any point on the parabola y=x2 and Q any point where normal line P intersects the parabola then the line segment PQ has the shortest possible length when a=12.

Expert Solution

Explanation of Solution

Use the information deduced in part (a).

If P(a,a2) is the given point, then the point Q is given by Q(a12a,(a12a)2).

Let the distance between these two points is DPQ. Then,

DPQ2=((2a+12a)2+(a2(a+12a)2)2)2=4a2+2+14a2+(a2(a2+1+14a2))2=4a2+2+14a2+1+12a2+116a4

Differentiate the above equation with respect to a.

2DPQdDPQda=8a12a31a314a5

Equate the above equation to 0.

8a12a31a314a5=032a62a24a214a5=032a66a21=0

On simplifying the above equation, the only possible value of a is a=12.

Compute the second derivative of DPQ with respect to a as follows.

d2DPQda2=8+92a4+54a6

As d2DPQda2>0, this is the minimum distance between P and Q.

That is the segment PQ has the shortest possible length when a=12.

Hence the proof.

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