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 3, Problem 17P

(a)

To determine

To show: If two lines intersect at an angle α then tanα=m2m11+m1m2.

Expert Solution

Explanation of Solution

Given:

Two lines are L1 and L2.

The slope of the lines L1 and L2 are m1 and m2 respectively.

Calculation:

Let ϕ1 and ϕ2 be an inclination angle of the line L1 and L2 respectively.

Here, the slopes of the line L1 and L2 are tanϕ1 and tanϕ2.

The angle between two lines are α=|ϕ2ϕ1|.

If the angle ϕ2>ϕ1,

α=ϕ2ϕ1

Take tangent on both sides,

tanα=tan(ϕ2ϕ1)

Since the identity tan(xy)=tanxtany1+tanxtany,

tanα=tanϕ2tanϕ11+tanϕ2tanϕ1=m2m11+m1m2

Hence the required result is proved.

(b)

To determine

To find: The angle between each pair of curves at each points.

Expert Solution

Answer to Problem 17P

The angle between two curves y=x2 and y=(x2)2 are, 53° and 127° and

the angle between two curves x2y2=3 and x24x+y2+3=0 are, α=117° and 63°.

Explanation of Solution

Given:

The pair of equation are

(i) y=x2 and y=(x2)2.

(ii) x2-y2=3 and x24x+y2+3=0.

Calculation:

Obtain the intersection point of y=x2 and y=(x2)2.

x2=(x2)2x2=x2+44x0=4(x1)x=1

Thus, the intersection is (1,1).

The slope of tangent line to the curve y=x2  is, dydx=2x.

Thus, the slope of tangent to the cure y=x2  at (1,1) is, m1=2.

The slope of the line  y=(x2)2 at (1,1) is,

dydx=ddx(x2)2=2(x2)ddx(x2)=2(x2)(10)=2x4 .

Substitute (1,1) in m2=dydx,

m2=2(1)4=2

Thus, the slope of the tangent to the curve at (1,1) is, m2=2.

Form part (a), tanα=m2m11+m1m2 where α is angle between two lines,

tanα=221+(2)(2)=43=43α=tan1(43)

The value α=53°.

The another angle is 180α, 127°.

Therefore, the angle between two curves y=x2 and y=(x2)2 are, 53° and 127°.

(ii)

Obtain the intersection point of x2y2=3 and x24x+y2+3=0.

x24x+y2+3=0

Substitute y2=x23,

x24x+x23+3=02x24x=02x(x2)=0x=0 or x=2

Here, x=0 is extraneous.

Substitute x=2 in y2=x23 and obtain y.

y2=223y2=1y=±1

Thus, the points of intersection are, (2,1) and (2,1).

The slope of tangent line to the curve x2y2=3 is computed as follows,

ddx(x2y2)=ddx(3)2x2ydydx=02ydydx=2xdydx=xy

Therefore, the slope of the tangent is, m1=xy

Thus, the slope of tangent to the cure x2y2=3 at the point (2,1) is, m1=2 and at the point (2,1) is m2=2.

The slope of the line x24x+y2+3=0 is computed as follows,

ddx(x24x+y2+3=0)=0ddx(x2)4ddx(x)+ddx(y2)+ddx(3)=02x4(1)+2ydydx+0=02ydydx=2x+4

That is, dydx=2x+42y.

Substitute (2,1) in dydx=2x+42y,

m3=2(2)+42(1)=0

Substitute (2,1) in dydx=2x+42y,

m4=2(2)+42(1)=0

Thus, the slope of the tangent to the curve x24x+y2+3=0 at the point (2,1) is, m3=0 and at the point (2,1) is m4=0.

At the point (2,1),

The angle between two lines at tanα=m3m11+m1m3 .

Substitute m1=2 and m3=0,

tanα=021+(2)0=2α=tan1(2)=117°

The value α=117° and another angle is 180α=63°.

At the point (2,1),

The angle between two lines is ,tanβ=m4m21+m2m4.

Substitute m2=2 and m4=0,

tanβ=0(2)1+(2)0=2β=tan1(2)=63°

The value β=63° and another angle is 18063°=117°.

Therefore, the angle between two curves x2y2=3 and  x24x+y2+3=0 are, 117° and 63°.

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