🧮

Complete Calculus Exercises

All 10 Problems with Working Visualizations!

Find the slope of the tangent

(i) \( y = x^4 + 2x^2 - x \) at \( x = 1 \)

Step 1: Find the derivative of the function to get the slope function. \[ \frac{dy}{dx} = \frac{d}{dx}(x^4) + \frac{d}{dx}(2x^2) - \frac{d}{dx}(x) \]

Step 2: Apply the power rule to each term. \[ \frac{dy}{dx} = 4x^3 + 4x - 1 \]

Step 3: Evaluate the derivative at \( x = 1 \) to find the slope. \[ \left.\frac{dy}{dx}\right|_{x=1} = 4(1)^3 + 4(1) - 1 = 4 + 4 - 1 = 7 \]

Click "Show Graph" to visualize

(ii) \( x = a \cos^3 t, y = b \sin^3 t \) at \( t = \frac{\pi}{2} \)

Step 1: Find dx/dt and dy/dt. \[ \frac{dx}{dt} = -3a \cos^2 t \sin t \] \[ \frac{dy}{dt} = 3b \sin^2 t \cos t \]

Step 2: The slope dy/dx is (dy/dt)/(dx/dt). \[ \frac{dy}{dx} = \frac{3b \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{b}{a} \tan t \]

Step 3: Evaluate at \( t = \frac{\pi}{2} \). \[ \left.\frac{dy}{dx}\right|_{t=\frac{\pi}{2}} = -\frac{b}{a} \tan \left(\frac{\pi}{2}\right) \] The tangent is undefined (vertical tangent) since tan(π/2) is undefined.

Click "Show Graph" to visualize

Find point where tangent is parallel

Find the point on the curve \( y = x^2 - 5x + 4 \) at which the tangent is parallel to the line \( 3x + y = 7 \).

Step 1: Find the slope of the given line. \[ 3x + y = 7 \Rightarrow y = -3x + 7 \] Slope (m) = -3

Step 2: Find the derivative of the curve. \[ \frac{dy}{dx} = 2x - 5 \]

Step 3: Set the derivative equal to the line's slope. \[ 2x - 5 = -3 \] \[ 2x = 2 \] \[ x = 1 \]

Step 4: Find the y-coordinate when x=1. \[ y = (1)^2 - 5(1) + 4 = 1 - 5 + 4 = 0 \] The point is (1, 0).

Click "Show Graph" to visualize

Find points where normal is parallel

Find the points on the curve \( y = x^3 - 6x^2 + x + 3 \) where the normal is parallel to the line \( x + y = 1729 \).

Step 1: Find slope of given line. \[ x + y = 1729 \Rightarrow y = -x + 1729 \] Slope (m) = -1

Step 2: Since normal is parallel to this line, normal's slope = -1. Therefore, tangent's slope = -1/(-1) = 1 (since normal is perpendicular to tangent).

Step 3: Find derivative of curve. \[ \frac{dy}{dx} = 3x^2 - 12x + 1 \] Set equal to 1 (tangent slope): \[ 3x^2 - 12x + 1 = 1 \] \[ 3x^2 - 12x = 0 \] \[ 3x(x - 4) = 0 \] \[ x = 0 \text{ or } x = 4 \]

Step 4: Find corresponding y-values.

For x = 0:

\[ y = 0 - 0 + 0 + 3 = 3 \] Point: (0, 3)

For x = 4:

\[ y = 64 - 96 + 4 + 3 = -25 \] Point: (4, -25)

Click "Show Graph" to visualize

Find points with horizontal tangent

Find the points on the curve \( y^2 - 4xy = x^2 + 5 \) for which the tangent is horizontal.

Step 1: Differentiate implicitly with respect to x. \[ 2y \frac{dy}{dx} - 4\left(y + x \frac{dy}{dx}\right) = 2x \]

Step 2: Collect terms with dy/dx. \[ 2y \frac{dy}{dx} - 4x \frac{dy}{dx} = 2x + 4y \] \[ \frac{dy}{dx}(2y - 4x) = 2x + 4y \] \[ \frac{dy}{dx} = \frac{2x + 4y}{2y - 4x} = \frac{x + 2y}{y - 2x} \]

Step 3: For horizontal tangent, dy/dx = 0. \[ \frac{x + 2y}{y - 2x} = 0 \Rightarrow x + 2y = 0 \Rightarrow x = -2y \]

Step 4: Substitute x = -2y into original equation. \[ y^2 - 4(-2y)(y) = (-2y)^2 + 5 \] \[ y^2 + 8y^2 = 4y^2 + 5 \] \[ 9y^2 = 4y^2 + 5 \] \[ 5y^2 = 5 \] \[ y^2 = 1 \] \[ y = 1 \text{ or } y = -1 \] Corresponding x-values: For y = 1: x = -2 For y = -1: x = 2 Points: (-2, 1) and (2, -1)

Click "Show Graph" to visualize

Find tangent and normal lines

(i) \( y = x^2 - x^4 \) at (1, 0)

Step 1: Find the derivative. \[ \frac{dy}{dx} = 2x - 4x^3 \]

Step 2: Evaluate at x = 1. \[ \left.\frac{dy}{dx}\right|_{x=1} = 2(1) - 4(1)^3 = 2 - 4 = -2 \] Slope of tangent (m) = -2

Step 3: Equation of tangent line at (1, 0): \[ y - 0 = -2(x - 1) \] \[ y = -2x + 2 \]

Step 4: Slope of normal line = -1/m = 1/2 Equation of normal line: \[ y - 0 = \frac{1}{2}(x - 1) \] \[ y = \frac{1}{2}x - \frac{1}{2} \]

Click "Show Graph" to visualize

(ii) \( y = x^4 + 2e^x \) at (0, 2)

Step 1: Find the derivative. \[ \frac{dy}{dx} = 4x^3 + 2e^x \]

Step 2: Evaluate at x = 0. \[ \left.\frac{dy}{dx}\right|_{x=0} = 0 + 2e^0 = 2 \] Slope of tangent (m) = 2

Step 3: Equation of tangent line at (0, 2): \[ y - 2 = 2(x - 0) \] \[ y = 2x + 2 \]

Step 4: Slope of normal line = -1/m = -1/2 Equation of normal line: \[ y - 2 = -\frac{1}{2}(x - 0) \] \[ y = -\frac{1}{2}x + 2 \]

Click "Show Graph" to visualize

Find tangent and normal lines (continued)

(iii) \( y = x \sin x \) at \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \)

Step 1: Find the derivative using product rule. \[ \frac{dy}{dx} = \sin x + x \cos x \]

Step 2: Evaluate at \( x = \frac{\pi}{2} \). \[ \left.\frac{dy}{dx}\right|_{x=\pi/2} = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2}\cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \] Slope of tangent (m) = 1

Step 3: Equation of tangent line: \[ y - \frac{\pi}{2} = 1\left(x - \frac{\pi}{2}\right) \] \[ y = x \]

Step 4: Slope of normal line = -1/m = -1 Equation of normal line: \[ y - \frac{\pi}{2} = -1\left(x - \frac{\pi}{2}\right) \] \[ y = -x + \pi \]

Click "Show Graph" to visualize

(iv) \( x = \cos t, y = 2 \sin^2 t \) at \( t = \frac{\pi}{3} \)

Step 1: Find dx/dt and dy/dt. \[ \frac{dx}{dt} = -\sin t \] \[ \frac{dy}{dt} = 4 \sin t \cos t \]

Step 2: Find dy/dx = (dy/dt)/(dx/dt). \[ \frac{dy}{dx} = \frac{4 \sin t \cos t}{-\sin t} = -4 \cos t \]

Step 3: Evaluate at \( t = \frac{\pi}{3} \). \[ \left.\frac{dy}{dx}\right|_{t=\pi/3} = -4 \cos\left(\frac{\pi}{3}\right) = -4 \times \frac{1}{2} = -2 \] Slope of tangent (m) = -2

Step 4: Find point coordinates at \( t = \frac{\pi}{3} \). \[ x = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] \[ y = 2 \sin^2\left(\frac{\pi}{3}\right) = 2 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{2} \] Point: \(\left(\frac{1}{2}, \frac{3}{2}\right)\)

Step 5: Equation of tangent line: \[ y - \frac{3}{2} = -2\left(x - \frac{1}{2}\right) \] \[ y = -2x + \frac{1}{2} + \frac{3}{2} \] \[ y = -2x + 2 \]

Step 6: Slope of normal line = -1/m = 1/2 Equation of normal line: \[ y - \frac{3}{2} = \frac{1}{2}\left(x - \frac{1}{2}\right) \] \[ y = \frac{1}{2}x - \frac{1}{4} + \frac{3}{2} \] \[ y = \frac{1}{2}x + \frac{5}{4} \]

Click "Show Graph" to visualize

Find orthogonal tangents

Find the equations of the tangents to the curve \( y = 1 + x^3 \) for which the tangent is orthogonal with the line \( x + 12y = 12 \).

Step 1: Find slope of given line. \[ x + 12y = 12 \Rightarrow y = -\frac{1}{12}x + 1 \] Slope (m) = -1/12

Step 2: For orthogonal lines, product of slopes = -1. So required tangent slope (m_t) satisfies: \[ m_t \times \left(-\frac{1}{12}\right) = -1 \] \[ m_t = 12 \]

Step 3: Find derivative of curve. \[ \frac{dy}{dx} = 3x^2 \] Set equal to 12: \[ 3x^2 = 12 \] \[ x^2 = 4 \] \[ x = 2 \text{ or } x = -2 \]

Step 4: Find points and tangent equations.

For x = 2:

\[ y = 1 + 8 = 9 \] Tangent equation: \[ y - 9 = 12(x - 2) \] \[ y = 12x - 15 \]

For x = -2:

\[ y = 1 - 8 = -7 \] Tangent equation: \[ y - (-7) = 12(x - (-2)) \] \[ y = 12x + 17 \]

Click "Show Graph" to visualize

Find parallel tangents

Find the equations of the tangents to the curve \( y = \frac{x + 1}{x - 1} \) which are parallel to the line \( x + 2y = 6 \).

Step 1: Find slope of given line. \[ x + 2y = 6 \Rightarrow y = -\frac{1}{2}x + 3 \] Slope (m) = -1/2

Step 2: Find derivative of curve using quotient rule. \[ \frac{dy}{dx} = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{-2}{(x-1)^2} \]

Step 3: Set derivative equal to -1/2. \[ \frac{-2}{(x-1)^2} = -\frac{1}{2} \] \[ \frac{2}{(x-1)^2} = \frac{1}{2} \] \[ (x-1)^2 = 4 \] \[ x-1 = 2 \text{ or } x-1 = -2 \] \[ x = 3 \text{ or } x = -1 \]

Step 4: Find points and tangent equations.

For x = 3:

\[ y = \frac{4}{2} = 2 \] Tangent equation: \[ y - 2 = -\frac{1}{2}(x - 3) \] \[ y = -\frac{1}{2}x + \frac{7}{2} \]

For x = -1:

\[ y = \frac{0}{-2} = 0 \] Tangent equation: \[ y - 0 = -\frac{1}{2}(x - (-1)) \] \[ y = -\frac{1}{2}x - \frac{1}{2} \]

Click "Show Graph" to visualize

Find tangent and normal to parametric curve

Find the equation of tangent and normal to the curve given by \( x = 7 \cos t \) and \( y = 2 \sin t, t \in \mathbb{R} \) at any point on the curve.

Step 1: Find dx/dt and dy/dt. \[ \frac{dx}{dt} = -7 \sin t \] \[ \frac{dy}{dt} = 2 \cos t \]

Step 2: Find dy/dx = (dy/dt)/(dx/dt). \[ \frac{dy}{dx} = \frac{2 \cos t}{-7 \sin t} = -\frac{2}{7} \cot t \]

Step 3: General point on curve: \( (7 \cos t, 2 \sin t) \) Tangent line equation: \[ y - 2 \sin t = -\frac{2}{7} \cot t (x - 7 \cos t) \]

Step 4: Normal line has slope = -1/(dy/dx) = (7/2) tan t Normal line equation: \[ y - 2 \sin t = \frac{7}{2} \tan t (x - 7 \cos t) \]

Click "Show Graph" to visualize

Find angle between curves

Find the angle between the rectangular hyperbola \( xy = 2 \) and the parabola \( x^2 + 4y = 0 \).

Step 1: Find intersection points. From \( x^2 + 4y = 0 \), \( y = -x^2/4 \). Substitute into \( xy = 2 \): \[ x(-x^2/4) = 2 \] \[ -x^3/4 = 2 \] \[ x^3 = -8 \] \[ x = -2 \] Then \( y = -(-2)^2/4 = -1 \) Intersection point: (-2, -1)

Step 2: Find slope of hyperbola \( xy = 2 \) at (-2, -1). Differentiate implicitly: \[ y + x \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{y}{x} \] At (-2, -1): \[ m_1 = -\frac{-1}{-2} = -\frac{1}{2} \]

Step 3: Find slope of parabola \( x^2 + 4y = 0 \) at (-2, -1). Differentiate implicitly: \[ 2x + 4 \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{x}{2} \] At (-2, -1): \[ m_2 = -\frac{-2}{2} = 1 \]

Step 4: Find angle θ between curves using: \[ \tan θ = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \] \[ \tan θ = \left| \frac{1 - (-0.5)}{1 + (-0.5)(1)} \right| = \left| \frac{1.5}{0.5} \right| = 3 \] \[ θ = \tan^{-1}(3) \approx 71.57^\circ \]

Click "Show Graph" to visualize

Show curves cut orthogonally

Show that the two curves \( x^2 - y^2 = r^2 \) and \( xy = c^2 \) where \( c, r \) are constants, cut orthogonally.

Step 1: Find dy/dx for both curves.

For \( x^2 - y^2 = r^2 \):

\[ 2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = m_1 \]

For \( xy = c^2 \):

\[ y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = m_2 \]

Step 2: At intersection points, show \( m_1 \times m_2 = -1 \). \[ m_1 \times m_2 = \left(\frac{x}{y}\right) \times \left(-\frac{y}{x}\right) = -1 \] This shows the tangents are perpendicular, so the curves cut orthogonally.

Click "Show Graph" to visualize