Problem Solving: Finding the constant of integration

Integration and Differentiation Exercises

1. Given that the gradient of a curve is \( \displaystyle 2x^2 + 7x \) and that the curve passes through the origin, determine the equation of the curve.

Solution

Let the equation of the curve be \( \displaystyle y = f(x) \).

The gradient of the curve is given by:

\( \displaystyle \frac{dy}{dx} = 2x^2 + 7x \)

Integrating with respect to \( \displaystyle x \) to find \( \displaystyle y \):

\( \begin{aligned} y &= \int (2x^2 + 7x) \, dx \\ y &= \frac{2x^3}{3} + \frac{7x^2}{2} + C \end{aligned} \)

Since the curve passes through the origin \( \displaystyle (0, 0) \), substitute \( \displaystyle x = 0 \) and \( \displaystyle y = 0 \):

\( \begin{aligned} 0 &= \frac{2(0)^3}{3} + \frac{7(0)^2}{2} + C \\ C &= 0 \end{aligned} \)

Therefore, the equation of the curve is:

\( \displaystyle y = \frac{2}{3}x^3 + \frac{7}{2}x^2 \)

2. A curve is such that \( \displaystyle \frac{dy}{dx} = 6x + k \), where \( \displaystyle k \) is a constant. The gradient of the normal to the curve at the point \( \displaystyle (1, -3) \) is \( \displaystyle \frac{1}{2} \). Find the equation of the curve.

Solution

The gradient of the normal at \( \displaystyle x = 1 \) is \( \displaystyle \frac{1}{2} \). Since the product of gradients of the normal and the tangent is \( \displaystyle -1 \), the gradient of the tangent at \( \displaystyle x = 1 \) is \( \displaystyle -2 \).

Substitute \( \displaystyle x = 1 \) and \( \displaystyle \frac{dy}{dx} = -2 \) into the given derivative equation:

\( \begin{aligned} 6(1) + k &= -2 \\ 6 + k &= -2 \\ k &= -8 \end{aligned} \)

So, the gradient function is \( \displaystyle \frac{dy}{dx} = 6x - 8 \).

Integrating to find the curve's equation:

\( \begin{aligned} y &= \int (6x - 8) \, dx \\ y &= 3x^2 - 8x + C \end{aligned} \)

Given the curve passes through \( \displaystyle (1, -3) \), substitute \( \displaystyle x = 1 \) and \( \displaystyle y = -3 \):

\( \begin{aligned} -3 &= 3(1)^2 - 8(1) + C \\ -3 &= 3 - 8 + C \\ -3 &= -5 + C \\ C &= 2 \end{aligned} \)

Therefore, the equation of the curve is:

\( \displaystyle y = 3x^2 - 8x + 2 \)

3. The curve \( \displaystyle C \), with equation \( \displaystyle y = f(x) \), passes through the point \( \displaystyle (1, 2) \) and \( \displaystyle f'(x) = 2x^3 - \frac{1}{x^2} \). Find the equation of \( \displaystyle C \) in the form \( \displaystyle y = f(x) \).

Solution

We are given \( \displaystyle f'(x) = 2x^3 - x^{-2} \).

Integrating to find \( \displaystyle f(x) \):

\( \begin{aligned} f(x) &= \int \left( 2x^3 - x^{-2} \right) dx \\ f(x) &= \frac{2x^4}{4} - \frac{x^{-1}}{-1} + C \\ f(x) &= \frac{1}{2}x^4 + \frac{1}{x} + C \end{aligned} \)

Since the curve passes through \( \displaystyle (1, 2) \), substitute \( \displaystyle x = 1 \) and \( \displaystyle f(1) = 2 \):

\( \begin{aligned} 2 &= \frac{1}{2}(1)^4 + \frac{1}{1} + C \\ 2 &= 1.5 + C \\ C &= 0.5 = \frac{1}{2} \end{aligned} \)

Therefore, the equation of the curve is:

\( \displaystyle y = \frac{1}{2}x^4 + \frac{1}{x} + \frac{1}{2} \)

4. The gradient of a particular curve is given by \( \displaystyle \frac{dy}{dx} = \frac{\sqrt{x} + 3}{x^2} \). Given that the curve passes through the point \( \displaystyle (9, 0) \), find an equation of the curve.

Solution

First, express the gradient function in index form:

\( \begin{aligned} \frac{dy}{dx} &= \frac{x^{\frac{1}{2}} + 3}{x^2} \\ &= x^{-\frac{3}{2}} + 3x^{-2} \end{aligned} \)

Integrating to find the curve's equation:

\( \begin{aligned} y &= \int \left( x^{-\frac{3}{2}} + 3x^{-2} \right) dx \\ y &= \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + \frac{3x^{-1}}{-1} + C \\ y &= -2x^{-\frac{1}{2}} - 3x^{-1} + C \\ y &= -\frac{2}{\sqrt{x}} - \frac{3}{x} + C \end{aligned} \)

The curve passes through \( \displaystyle (9, 0) \), so substitute \( \displaystyle x = 9 \) and \( \displaystyle y = 0 \):

\( \begin{aligned} 0 &= -\frac{2}{\sqrt{9}} - \frac{3}{9} + C \\ 0 &= -\frac{2}{3} - \frac{1}{3} + C \\ 0 &= -1 + C \\ C &= 1 \end{aligned} \)

Therefore, the equation of the curve is:

\( \displaystyle y = -\frac{2}{\sqrt{x}} - \frac{3}{x} + 1 \)

5. The curve with equation \( \displaystyle y = f(x) \) passes through the point \( \displaystyle (-1, 0) \). Given that \( \displaystyle f'(x) = 9x^2 + 4x - 3 \), find \( \displaystyle f(x) \).

Solution

Integrating the derivative to find \( \displaystyle f(x) \):

\( \begin{aligned} f(x) &= \int (9x^2 + 4x - 3) \, dx \\ f(x) &= \frac{9x^3}{3} + \frac{4x^2}{2} - 3x + C \\ f(x) &= 3x^3 + 2x^2 - 3x + C \end{aligned} \)

Since the curve passes through \( \displaystyle (-1, 0) \), substitute \( \displaystyle x = -1 \) and \( \displaystyle f(-1) = 0 \):

\( \begin{aligned} 0 &= 3(-1)^3 + 2(-1)^2 - 3(-1) + C \\ 0 &= -3 + 2 + 3 + C \\ 0 &= 2 + C \\ C &= -2 \end{aligned} \)

Therefore, the function is:

\( \displaystyle f(x) = 3x^3 + 2x^2 - 3x - 2 \)

6. \( \displaystyle \frac{dy}{dx} = 3x^{-\frac{1}{2}} - 2x\sqrt{x}, x > 0 \). Given that \( \displaystyle y = 10 \) at \( \displaystyle x = 4 \), find \( \displaystyle y \) in terms of \( \displaystyle x \), giving each term in its simplest form.

Solution

First, express the gradient function entirely in index form. Note that \( \displaystyle 2x\sqrt{x} = 2x^{\frac{3}{2}} \):

\( \displaystyle \frac{dy}{dx} = 3x^{-\frac{1}{2}} - 2x^{\frac{3}{2}} \)

Integrating with respect to \( \displaystyle x \):

\( \begin{aligned} y &= \int \left( 3x^{-\frac{1}{2}} - 2x^{\frac{3}{2}} \right) dx \\ y &= \frac{3x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{2x^{\frac{5}{2}}}{\frac{5}{2}} + C \\ y &= 6x^{\frac{1}{2}} - \frac{4}{5}x^{\frac{5}{2}} + C \end{aligned} \)

Substitute \( \displaystyle x = 4 \) and \( \displaystyle y = 10 \):

\( \begin{aligned} 10 &= 6(4)^{\frac{1}{2}} - \frac{4}{5}(4)^{\frac{5}{2}} + C \\ 10 &= 6(2) - \frac{4}{5}(32) + C \\ 10 &= 12 - \frac{128}{5} + C \\ 10 &= 12 - 25.6 + C \\ 10 &= -13.6 + C \\ C &= 23.6 = \frac{118}{5} \end{aligned} \)

Therefore, the equation in its simplest form is:

\( \displaystyle y = 6x^{\frac{1}{2}} - \frac{4}{5}x^{\frac{5}{2}} + \frac{118}{5} \)

7. A curve \( \displaystyle y = f(x) \) has a stationary point at \( \displaystyle P(2, -13) \) and is such that \( \displaystyle f'(x) = 2x^2 + 3x - k \), where \( \displaystyle k \) is a constant.

(a) Find the \( \displaystyle x \)-coordinate of the other stationary point, \( \displaystyle Q \), on the curve \( \displaystyle y = f(x) \).

(b) Determine the nature of each of the stationary points \( \displaystyle P \) and \( \displaystyle Q \).

Solution

(a) At the stationary point \( \displaystyle P(2, -13) \), the gradient \( \displaystyle f'(x) \) is zero. Substitute \( \displaystyle x = 2 \) into \( \displaystyle f'(x) = 0 \):

\( \begin{aligned} f'(2) &= 2(2)^2 + 3(2) - k = 0 \\ 2(4) + 6 - k &= 0 \\ 8 + 6 - k &= 0 \\ 14 - k &= 0 \\ k &= 14 \end{aligned} \)

Now, rewrite the full expression for \( \displaystyle f'(x) \) and equate to zero to find the other stationary point:

\( \begin{aligned} 2x^2 + 3x - 14 &= 0 \\ (2x + 7)(x - 2) &= 0 \end{aligned} \)

This gives \( \displaystyle x = 2 \) (which is point \( \displaystyle P \)) or \( \displaystyle 2x + 7 = 0 \implies x = -\frac{7}{2} \).

The \( \displaystyle x \)-coordinate of the other stationary point \( \displaystyle Q \) is \( \displaystyle -\frac{7}{2} \) (or \( \displaystyle -3.5 \)).

(b) Find the second derivative \( \displaystyle f''(x) \):

\( \displaystyle f''(x) = \frac{d}{dx}(2x^2 + 3x - 14) = 4x + 3 \)

Check the nature at point \( \displaystyle P \) (\( \displaystyle x = 2 \)):

\( \displaystyle f''(2) = 4(2) + 3 = 11 \)

Since \( \displaystyle 11 > 0 \), point \( \displaystyle P \) is a minimum.

Check the nature at point \( \displaystyle Q \) (\( \displaystyle x = -3.5 \)):

\( \displaystyle f''(-3.5) = 4(-3.5) + 3 = -14 + 3 = -11 \)

Since \( \displaystyle -11 < 0 \), point \( \displaystyle Q \) is a maximum.

8. A curve is such that \( \displaystyle \frac{dy}{dx} = 5x\sqrt{x} + 2 \). Given that the curve passes through the point \( \displaystyle (1, 3) \), find:

(a) the equation of the curve

(b) the equation of the tangent to the curve when \( \displaystyle x = 4 \).

Solution

(a) Write the gradient function in index form:

\( \displaystyle \frac{dy}{dx} = 5x^{\frac{3}{2}} + 2 \)

Integrating to find \( \displaystyle y \):

\( \begin{aligned} y &= \int \left( 5x^{\frac{3}{2}} + 2 \right) dx \\ y &= \frac{5x^{\frac{5}{2}}}{\frac{5}{2}} + 2x + C \\ y &= 2x^{\frac{5}{2}} + 2x + C \end{aligned} \)

The curve passes through \( \displaystyle (1, 3) \):

\( \begin{aligned} 3 &= 2(1)^{\frac{5}{2}} + 2(1) + C \\ 3 &= 2 + 2 + C \\ C &= -1 \end{aligned} \)

Therefore, the equation of the curve is:

\( \displaystyle y = 2x^{\frac{5}{2}} + 2x - 1 \)

(b) First, find the \( \displaystyle y \)-coordinate and the gradient when \( \displaystyle x = 4 \).

Substitute \( \displaystyle x = 4 \) into the curve equation to find \( \displaystyle y \):

\( \begin{aligned} y &= 2(4)^{\frac{5}{2}} + 2(4) - 1 \\ y &= 2(32) + 8 - 1 \\ y &= 64 + 7 = 71 \end{aligned} \)

So the point is \( \displaystyle (4, 71) \).

Now substitute \( \displaystyle x = 4 \) into \( \displaystyle \frac{dy}{dx} \) to find the gradient \( \displaystyle m \):

\( \begin{aligned} m &= 5(4)^{\frac{3}{2}} + 2 \\ m &= 5(8) + 2 \\ m &= 40 + 2 = 42 \end{aligned} \)

The equation of the tangent using \( \displaystyle y - y_1 = m(x - x_1) \) is:

\( \begin{aligned} y - 71 &= 42(x - 4) \\ y &= 42x - 168 + 71 \\ y &= 42x - 97 \end{aligned} \)

9. The displacement of a particle at time \( \displaystyle t \) is given by the function \( \displaystyle f(t) \), where \( \displaystyle f(0) = 0 \). Given that the velocity of the particle is given by \( \displaystyle f'(t) = 10 - 5t \),

(a) find \( \displaystyle f(t) \)

(b) determine the displacement of the particle when \( \displaystyle t = 3 \).

Solution

(a) To find displacement \( \displaystyle f(t) \), integrate velocity \( \displaystyle f'(t) \):

\( \begin{aligned} f(t) &= \int (10 - 5t) \, dt \\ f(t) &= 10t - \frac{5}{2}t^2 + C \end{aligned} \)

Given \( \displaystyle f(0) = 0 \):

\( \begin{aligned} 0 &= 10(0) - \frac{5}{2}(0)^2 + C \implies C = 0 \end{aligned} \)

Therefore, the displacement function is:

\( \displaystyle f(t) = 10t - 2.5t^2 \)

(b) Substitute \( \displaystyle t = 3 \) into the displacement function:

\( \begin{aligned} f(3) &= 10(3) - 2.5(3)^2 \\ &= 30 - 2.5(9) \\ &= 30 - 22.5 \\ &= 7.5 \end{aligned} \)

The displacement of the particle when \( \displaystyle t = 3 \) is \( \displaystyle 7.5 \).

10. The height, in metres, of an arrow fired horizontally from the top of a castle is modelled by the function \( \displaystyle f(t) \), where \( \displaystyle f(0) = 35 \). Given that \( \displaystyle f'(t) = -9.8t \),

(a) find \( \displaystyle f(t) \).

(b) determine the height of the arrow when \( \displaystyle t = 1.5 \).

(c) write down the height of the castle according to this model.

(d) estimate the time it will take the arrow to hit the ground.

(e) state one assumption used in your calculation.

Solution

(a) Integrate \( \displaystyle f'(t) \) to find \( \displaystyle f(t) \):

\( \begin{aligned} f(t) &= \int -9.8t \, dt \\ f(t) &= -4.9t^2 + C \end{aligned} \)

Given \( \displaystyle f(0) = 35 \):

\( \displaystyle 35 = -4.9(0)^2 + C \implies C = 35 \)

Therefore, the height function is:

\( \displaystyle f(t) = -4.9t^2 + 35 \)

(b) Height when \( \displaystyle t = 1.5 \):

\( \begin{aligned} f(1.5) &= -4.9(1.5)^2 + 35 \\ &= -4.9(2.25) + 35 \\ &= -11.025 + 35 \\ &= 23.975 \text{ m} \end{aligned} \)

(c) The height of the castle corresponds to the height of the arrow at \( \displaystyle t = 0 \), which is explicitly given as:

\( \displaystyle 35 \text{ m} \)

(d) The arrow hits the ground when its height \( \displaystyle f(t) = 0 \):

\( \begin{aligned} -4.9t^2 + 35 &= 0 \\ 4.9t^2 &= 35 \\ t^2 &= \frac{35}{4.9} \\ t^2 &= \frac{350}{49} = \frac{50}{7} \\ t &= \sqrt{\frac{50}{7}} \approx 2.67 \text{ seconds} \end{aligned} \)

(e) One assumption is that there is no air resistance acting on the arrow (or that the ground is completely horizontal and flat relative to the castle base).

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