SLOPES, TANGENTS, AND NORMALS - Further Applications of the Derivative - The Calculus Primer

The Calculus Primer (2011)

Part VI. Further Applications of the Derivative

Chapter 20. SLOPES, TANGENTS, AND NORMALS

6—1.Slope of a Curve. We have seen that for any curve, y = f(x), the slope m of the tangent to the curve at any particular point (x1,y1) is given by m = tan ø = images , where the symbol images indicates the particular value which the variable expression images takes when x = x1 and y = y1. The reader is urgently cautioned not to think of images as meaning the derivative of y1 with respect to x1; such an interpretation would be meaningless since both x1 and y1 are constants, not variables.

images

EXAMPLE 1.Find the slope of the tangent to y = x4 + 2x3 − 5x + 3 at the point where x = 2.

Solution.

images

EXAMPLE 2.Find the slope of the tangent to y2 = 12x + 13 at the point (3,7).

Solution. Differentiating:

images

EXAMPLE 3.Find the slope of the curve 2x2 + y2 = 4x + 42 at the points where x = 3.

Solution. If x1 = 3, then y1 = ±6. Differentiating:

images

EXAMPLE 4.Find the slope of the curve

x2y2 + x3 − 2xy4 − 6y = 0

at the point (0,0).

Solution. Differentiating:

images

6—2.The Angle between Two Intersecting Curves. The reader will recall that the angle between two intersecting curves is defined as the angle between the tangents to the two curves, respectively, at their point of intersection P(x1,y1). Since

images

it is merely necessary first to determine m1 and m2 by differentiation, then to substitute the values of x1 and y1 in each of the two derivatives so found, and then to substitute these values in the expression for tan θ.

images

EXAMPLE. Find the angle of intersection between the circle

x2 + y2 = 25

and the parabola 4y2 = 9x.

images

Solution. Solving these equations simultaneously we have:

x = 4, y = ±3.

Letx1 = 4,y1 = ±3.

Differentiating:

images

hence, arc tan images = arc tan (− 3.417) = 106°20′ (approx.)

EXERCISE 6—1

1. Find the slope of the curve

images

at any point.

2. Show that the tangent to the circle x2 + y2 + 2x + 4y = 0 at the origin is parallel to the line x + 2y = 10.

3. Find the slope of the curve

y = x(x3 + 7)

at the point where x = 1.

4. At what point on the curve y2 = 3x3 is the slope equal to images?

5. Find the angle of intersection between the curves

x2 + y2 + 2x − 3 = 0andx2 + y2 = 7.

6. At what points on the circle x2 + y2 = k2 is the slope of the tangent to the circle equal to – images ?

7. Find the angle of intersection of the two parabolas x2 = 4py and y2 = 4px.

8. Prove that for all values of k, the equation kx = ev has the same slope, that is, the slope is independent of the value of k.

6—3.Equations of Tangent and Normal. The equation of the tangent, ST, to a curve at a given point P1(x1,y1) is easily derived.

images

The slope of the tangent in question is given by images ; since the point of tangency lies on the tangent as well as on the curve, we may use the point-slope formula for the equation of a line having a given slope and passing through a given point. Thus:

images

The normal is the line perpendicular to a tangent at the point of contact. Hence the equation of the normal, P1N, given by

images

EXAMPLE 1.Find the equations of the tangent and the normal to the curve 4x2 + 9y2 = 25 at the point where x = 2 and y is positive.

Solution.

images

At x1 = 2, y1 = ±1; taking y1 positive:

images

Hence, the equation of the tangent at (2,1) is

y – 1 = – images (x – 2),or8x + 9y – 25 = 0

and the equation of the normal at (2,1) is

y – 1 = – images (x – 2),or9x + 8y – 10 = 0.

EXAMPLE 2.Find the equations of the tangent and the normal to the equation 2x3 = y2 at the point where x = 1 and y is negative.

Solution.

images

At x1 = 1, y1 = ±images; taking y1 negative:

images

Hence, the equation of the tangent is

images

The equation of the normal:

images

6—4.Length of Subtangent and Subnormal. Referring to the diagram in §6—3, the segment RM is called the subtangent of the point P1; the length of the segment MQ is called the subnormal of P1. The lengths of these segments, for any curve f(x,y) = 0, can be readily derived.

images

images

EXAMPLE 1.Find the lengths of the subtangent and the subnormal to the curve x2 = 8y + 4 when x = 6.

images

At x1 = 6, y1 = 4;

images

EXAMPLE 2.Find the lengths of the subtangent and the subnormal to x2 = 2y3 at x = 4.

images

At x1 = 4, y1 = 2; hence,

images

EXERCISE 6—2

1. Find the length of the subtangent to 3x2y2 = 12 at x = 4.

2. Find the equation of the tangent and the normal to the curve x3 = y + xy at the point where x = − 2.

3. Find the lengths of the subtangent and the subnormal to the curve y2 = 4px at any point (x,y).

4. Find the equation of the tangent, and the length of the subtangent, to the circle x2 + y2 = 25 at x = 3.

5. Find the equation of the tangent to the curve a2(xy) = x3 + x2y at the origin.

6. Find the length of the subtangent to the curve x = kyn when x = k.

7. Find the length of the subtangent to the curve y = kx.

8. Prove that the area of the triangle formed by the coordinate axes and any tangent drawn to the curve 2xy = a2 is a constant and equal to a2.