Dr. J's Maths.com
Where the techniques of Maths
are explained in simple terms.

Calculus - Differentiation - First derivative.
Gradients and equations of curves, tangents, etc - Test Yourself 1 - Solutions.


 

Remember that the first derivative is the gradient function of an original function.

So it is used to answer questions requiring information about the slope of a curve.

Gradients   1. y = 2x3 - 6x where gradient equals zero?

2. - gradient equals 3?

    3. At what value(s) of x does the curve

y = 2x2 (x - 1)4 have its gradient equal to zero?

 

 4. At what point on the curve

does the gradient equal -1?
    5. y = x3 + ax2 + bx - 10 cuts the x-axis at x = 2 and it has a gradient of zero at
x = -1.

6. Evaluate f '(1) if f(x) = 5x2(4x2 - 1)5.

    7. f(x) = x2 + x

f '(x) = 2x + 1

f"(x) = 2

∴ x2 + x = 2

x2 + x - 2 = 0

(x + 2)(x - 1) = 0

x = -2 or +1

y = 5 or 3

Points are (-2, 5) or (1, 3).

 8.
    9. Evaluate f '(0) if.

10. Prove that the tangents at the points
(2, -3) and (1, -3) on the curve y = x2 - 3x -1 are perpendicular to one another.

    11. y = ax2 + bx (i)

y' = 2ax + b (ii)

At the point (2, 1)

Sub into (i): 1 = 4a + 2b

Sub into (ii): -8 = 4a + b

Solve simultaneously - subtract (i) - (ii):

9 = b

a = -17/4 = -4.25

∴ a = -17/4 and b = 9.

 12. f(x) = ax2 + bx + c

f '(x) = 2ax + b

Substituting given values:

a = a + b + c so b = -c

1 = 2a + b

1 = 2a -c

∴ a = (1 + c)/2
Points of contact   13. The gradient of a tangent to the curve

y = x2 - 5x is -3.

Find the coordinates of the point of contact of the tangent to the curve .

y' = 2x - 5 = -3

x = 1 and so y = -4.

 14. The gradient of a tangent to the curve

y = x2 + 7x - 3 is 9.

Find the coordinates of the point of contact of the tangent to the curve.

y' = 2x + 7 = 9

x = 1 and so y = 5.
    15. y = x2 - 2x - 8 has gradient 8 and its equation is y = ax + b.

y' = 2x - 2 = 8

x = 5 and y = 7.

y - 7 = 8( x - 5)

y = 8x - 33

∴ a = 8 and b = -33.

16.
    17. For y = 2x - 5 the gradient is 2.

For y = x2 + bx + 4 the gradient function is y ' = 2x + b.

At x = 3, the gradients are the same. So:

2 = 6 + b

b = -4.

 18.
Equations of tangents & normals. Ordinary derivative. 19. Find the equation of the normal

to the curve y = 3x - 2x2 at x = 1.

y' = 3 - 4x

At x = 1, y = 1 and y' = -1

Gradient of normal = 1

Eqn is y - 1 = 1( x - 1)

y = x.

 20. Find the equation of the tangent to the curve at the point
where x = 1.

 
    21. The curve y = ax2 - 2x -14 has a gradient of 10 when x = 2. Find the value of a.

y' = 2ax - 2

x = 2 4a - 2 = 10

a = 3.

 22. Find the equation of the tangent to
y = 3x4 + 1 at the point (1, 4).

y' = 12x3

At x = 1, y' = 12
    23. (i)

y = x3 - 9x2 + 20x - 8

y ' = 3x2 - 18x + 20

At x = 1, y ' = 5.

Eqn of tangent is
y - 4 = 5(x - 1)

y = 5x - 1.

(ii) To be parallel,
gradient function = gradient of the line (i.e. -4)

3x2 - 18x + 20 = -4

3(x - 4)(x - 2) = 0

∴ x = 4 or x = 2

 24.
  Chain rule. 25.Find the equation of the tangent to

y = (2x - 3)2 at the point where x = 3.

y' = 2(2x - 3)× 2

At x = 3, y = 9 and y ' = 12

Eqn is y - 9 = 12(x - 3)

y = 12x - 25

 
 26. Find the equation of the normal to the curve f(x) = (2x - 1)2 at x = 2.

f '(x) = 2(2x - 1)× 2

= 4(2x - 1)

At x = 2, f (2) = 9 and f '(2) = 12
  Product rule. 27. y = x2 (3x - 2)3

 

 28.
    29.

30.
  Quotient rule. 31. 32.
Increasing/ decreasing functions.   33. For what values of x is the function f(x) = 6 - 3x - x2 increasing?

To be increasing, the function
must have a positive gradient.

f '(x) = -3 - 2x > 0

2x < -3

x < -1.5

 

 34. For what values of x is the function

f(x) = x3 + x2 - 5x - 6 decreasing?

f '(x) = 3x2 + 2x - 5

= (3x + 5) (x - 1)

x = -5/3 or x = 1

To be a decreasing function, f '(x) < 0

Testing on a number line:

f '(0) = -5 < 0

So -5/3 < x < 1

 
    35.

∴ function is monotonically decreasing for all x ≠ 1.5

36.

The function is not defined for x < 0 and the derivative is not defined at x = 0. So the function is decreasing for 0 < x < 0.488.
Applications to points of intersection, area, etc   37. y = 8 - 2x2

(i)

(ii) P and Q are where y = 0. So 8 - 2x2 = 0

x2 = 4 giving x = +2 or x = -2.

y ' = -4x

So at x = 2, y = 0 and y ' = -8

At x = -2, y = 0 and y' = +8

Using the negative reciprocals of these gradients in y - y1 = m(x - x1) give the two equations for the normals as:

At P: x - 8y - 2 = 0.

At Q: x + 8y + 2 = 0

(iii) To find R, solve simultaneously by adding: 2x = 0 giving x = 0 (i.e. the y axis).

Substituting back into the two equations gives y = -0.25.

So the coordinates of R are (0, -0.25) as required.

(iv) From the diagram, area = 0.5 × 4 × 0.25 = 0.5 u2.

 
    38. y = x + 3 and y = 5x - x2.

(i) For POI, x + 3 = 5x - x2

x2 - 4x + 3 = 0

(x - 3)(x - 1) = 0 giving x = 3 and x = 1

When x = 3, y = 6 and when x = 1, y = 4.

So P is (3, 6) and Q is (1, 4).

(ii) y ' = 5 - 2x

So gradient at P = 5 - 2(3) = -1 and at Q = 5 - 2(1) = 3.

∴ Eqn of tangent at P is x + y - 9 = 0 and eqn of tangent at Q is 3x - y + 1= 0.

(iii) At A, P meets the y axis at y = 9 and at B, Q meets the y axis at 1.

Distance between A and B = 9 - 1 = 8 units.

 
    39.