FIND THE SLOPE OF THE TANGENT LINE AT THE INDICATED POINT

Find the slope of the tangent of the following curves at the respective given points.

Problem 1 :

y = x⁴ + 2x² - x at x = 1

Solution :

Finding the first derivative,

dy/dx = 4x³ + 2(2x) - 1

dy/dx = 4x³ + 4x - 1

Applying x = 1 in the first derivative,

dy/dx at x = 1 ==> 4(1)³ + 4(1) - 1 

= 4 + 4 - 1

= 8 - 1 

= 7

So, slope of the tangent line at the given point is 7.

Problem 2 :

x = a cos³ t, y = b sin³ t at x = π/2

Solution :

x = a cos³ t, y = b sin³ t

(dy/dt) / (dx/dt) = 3b sin² t cos t / ( -3a cos² t sin t)

dy/dx = (-b/a) (sin t / cos t) 

Slope at x = π/2

dy/dx = (-b/a) (sin π/2 / cos π/2) 

= (-b/a) (1/0)

= undefined.

Problem 3 :

Find the point on the curve y = x² − 5x + 4 at which the tangent is parallel to the line 3x + y = 7

Solution :

y = x² − 5x + 4

dy/dx = 2x - 5(1) + 0

dy/dx = 2x - 5

The given line 3x + y = 7 is parallel to the tangent line drawn to the curve at the specific point.

So, slope of these two lines will be equal.

dy/dx = 2x - 5 -----(1)

3x + y = 7

y = -3x + 7

Slope = -3 -----(2)

(1) = (2)

2x - 5 = -3

2x = -3 + 5

2x = 2

x = 1

Applying x = 1 in y = x² − 5x + 4, we get

y = 1² − 5(1) + 4

y = 1 - 5 + 4

y = 0

So, the required point is (1, 0).

Problem 4 :

Find the points on the curve y = x³ − 6x² + x + 3 where the normal is parallel to the line x + y = 1729 .

Solution :

y = x³ − 6x² + x + 3

dy/dx = 3x² - 6(2x) + 1 + 0

= 3x² - 12x + 1 -----(1)

y = 1729 - x

Slope = -1

Slope of normal = -1/(-1)

= 1 ------(2)

(1) = (2)

3x² - 12x + 1 = 1

3x² - 12x + 1 - 1 = 0

3x² - 12x = 0

3x(x - 4) = 0

3x = 0 and x - 4 = 0

x = 0 and x = 4

Applying x = 0 in y = x³ − 6x² + x + 3

y = 0³ − 6(0)² + 0 + 3

y = 3

Applying x = 4 in y = x³ − 6x² + x + 3

y = 4³ − 6(4)² + 4 + 3

y = 64 - 96 + 4 + 3

y = 71 - 96

y = -25

So, the required points are (0, 3) and (4, -25).

Problem 5 :

Find the points on the curve y²- 4xy = x² + 5 for which the tangent is horizontal.

Solution :

Since the tangent line is horizontal, the slope of the horizontal line = 0.

2y (dy/dx) - 4[x (dy/dx) + y(1)] = 2x + 0

2y (dy/dx) - 4x (dy/dx) - 4y = 2x

(dy/dx) [2y - 4x] = 2x + 4y

(dy/dx) = (2x + 4y) / (2y - 4x)

Slope = 0

(2x + 4y) / (2y - 4x) = 0

2x + 4y = 0

4y = -2x

y = -x/2

Applying y = -x/2, we get

(-x/2)²- 4x(-x/2) = x² + 5

x²/4 + 2x² = x² + 5

9x² = 4x² + 20

9x² - 4x² = 20

5x² = 20

x² = 4

x = 2 and -2

When x = 2, y = -2/2 ==> -1

When x = -2, y = 2/2 ==> 1

So, the required points are (2, -1) and (-2, 1).