How to Find Where the Tangent Cuts the Curve Again
half dozen.4 Equation of a tangent to a bend (EMCH8)
At a given point on a curve, the slope of the curve is equal to the gradient of the tangent to the bend.
The derivative (or gradient function) describes the gradient of a curve at whatsoever point on the curve. Similarly, it also describes the gradient of a tangent to a bend at any point on the curve.
To decide the equation of a tangent to a bend:
- Find the derivative using the rules of differentiation.
- Substitute the \(x\)-coordinate of the given signal into the derivative to calculate the gradient of the tangent.
- Substitute the slope of the tangent and the coordinates of the given point into an appropriate form of the straight line equation.
- Make \(y\) the subject field of the formula.
The normal to a curve is the line perpendicular to the tangent to the curve at a given indicate.
\[m_{\text{tangent}} \times m_{\text{normal}} = -1\]Worked example 13: Finding the equation of a tangent to a bend
Find the equation of the tangent to the bend \(y=3{10}^{2}\) at the betoken \(\left(ane;3\right)\). Sketch the curve and the tangent.
Find the derivative
Use the rules of differentiation:
\brainstorm{align*} y &= 3{ten}^{2} \\ & \\ \therefore \frac{dy}{dx} &= iii \left( 2x \right) \\ &= 6x \end{marshal*}
Calculate the gradient of the tangent
To determine the slope of the tangent at the bespeak \(\left(1;3\right)\), we substitute the \(10\)-value into the equation for the derivative.
\begin{align*} \frac{dy}{dx} &= 6x \\ \therefore chiliad &= 6(1) \\ &= 6 \end{align*}
Make up one's mind the equation of the tangent
Substitute the gradient of the tangent and the coordinates of the given bespeak into the gradient-point form of the straight line equation.
\begin{align*} y-{y}_{ane} & = one thousand\left(10-{x}_{1}\correct) \\ y-three & = six\left(x-1\right) \\ y & = 6x-6+3 \\ y & = 6x-3 \finish{marshal*}
Sketch the bend and the tangent
Worked example fourteen: Finding the equation of a tangent to a curve
Given \(one thousand(x)= (x + 2)(2x + 1)^{2}\), decide the equation of the tangent to the bend at \(x = -i\) .
Determine the \(y\)-coordinate of the indicate
\begin{align*} one thousand(ten) &= (x + 2)(2x + 1)^{2} \\ thou(-i) &= (-1 + two)[ii(-one) + 1]^{2} \\ &= (1)(-one)^{ii} \\ & = i \end{align*}
Therefore the tangent to the curve passes through the point \((-ane;1)\).
Expand and simplify the given function
\begin{align*} g(ten) &= (x + 2)(2x + 1)^{ii} \\ &= (ten + ii)(4x^{ii} + 4x + 1) \\ &= 4x^{iii} + 4x^{ii} + x + 8x^{2} + 8x + 2 \\ &= 4x^{3} + 12x^{ii} + 9x + 2 \cease{marshal*}
Find the derivative
\begin{align*} 1000'(x) &= 4(3x^{2}) + 12(2x) + ix + 0 \\ &= 12x^{2} + 24x + ix \end{align*}
Calculate the gradient of the tangent
Substitute \(x = -\text{ane}\) into the equation for \(g'(ten)\):
\brainstorm{align*} thou'(-1) &= 12(-1)^{ii} + 24(-i) + 9 \\ \therefore grand &= 12 - 24 + 9 \\ &= -iii \end{align*}
Determine the equation of the tangent
Substitute the gradient of the tangent and the coordinates of the point into the gradient-point class of the straight line equation.
\begin{align*} y-{y}_{1} & = thou\left(ten-{10}_{1}\right) \\ y-1 & = -iii\left(10-(-1)\right) \\ y & = -3x - 3 + one \\ y & = -3x - two \cease{align*}
Worked case fifteen: Finding the equation of a normal to a curve
- Determine the equation of the normal to the bend \(xy = -4\) at \(\left(-1;4\right)\).
- Draw a rough sketch.
Discover the derivative
Make \(y\) the subject of the formula and differentiate with respect to \(x\):
\begin{align*} y &= -\frac{4}{x} \\ &= -4x^{-one} \\ & \\ \therefore \frac{dy}{dx} &= -four \left( -1x^{-two} \correct) \\ &= 4x^{-2} \\ &= \frac{four}{10^{2}} \end{align*}
Calculate the slope of the normal at \(\left(-1;4\correct)\)
First determine the gradient of the tangent at the given point:
\begin{marshal*} \frac{dy}{dx} &= \frac{4}{(-1)^{2}} \\ \therefore yard &= 4 \end{align*}
Use the gradient of the tangent to calculate the gradient of the normal:
\begin{align*} m_{\text{tangent}} \times m_{\text{normal}} &= -ane \\ 4 \times m_{\text{normal}} &= -1 \\ \therefore m_{\text{normal}} &= -\frac{1}{iv} \end{align*}
Find the equation of the normal
Substitute the slope of the normal and the coordinates of the given bespeak into the gradient-point form of the straight line equation.
\brainstorm{marshal*} y-{y}_{1} & = yard\left(x-{x}_{1}\right) \\ y-four & = -\frac{1}{4}\left(10-(-1)\right) \\ y & = -\frac{1}{4}ten - \frac{ane}{four} + iv\\ y & = -\frac{1}{4}x + \frac{15}{4} \end{align*}
Draw a rough sketch
Equation of a tangent to a curve
Textbook Exercise 6.5
Determine the equation of the tangent to the curve defined by \(F(x)=x^{3}+2x^{2}-7x+1\) at \(ten=2\).
\begin{align*} \text{Gradient of tangent }&= F'(10) \\ F'(x) &=3x^{2} +4x - seven \\ F'(two) &=3(2)^{2} + (4)(two) -7 \\ &=13 \\ \therefore \text{Tangent: } y &=13x +c \end{align*}
where \(c\) is the \(y\)-intercept.
Tangent meets \(F(x)\) at \((2;F(2))\)
\begin{marshal*} F(2) &=(2)^{3} + 2(2)^{ii} - 7(2) +one \\ &= 8 + eight -14 +1 \\ &=3 \\ \text{Tangent: } three &=thirteen(ii) + c \\ \therefore c &= - 23 \\ y & = 13x - 23 \stop{align*}
\(f(ten)=1-3x^{2}\) is equal to \(\text{5}\).
\begin{align*} \text{Slope of tangent } = f'(ten) = -6x \\ \therefore -6x &= 5 \\ \therefore x &= - \frac{5}{half dozen} \\ \text{And } f\left(- \frac{five}{six} \right) &=1-3 \left( - \frac{v}{6} \correct)^{two} \\ &=one-3 \left( \frac{25}{36} \correct) \\ &=1 - \frac{25}{12} \\ &= - \frac{xiii}{12} \\ \therefore & \left( - \frac{v}{6};- \frac{13}{12} \right) \cease{align*}
\(k(10)=\frac{1}{3}x^{two}+2x+1\) is equal to \(\text{0}\).
\begin{align*} \text{Gradient of tangent } = g'(10) = \frac{two}{iii}x+2 \\ \therefore \frac{2}{3}ten+2 &=0 \\ \frac{2}{3}10 &= -2\\ \therefore x&=-two \times \frac{three}{ii} \\ &=-3 \\ \text{And } one thousand(-iii) &= \frac{1}{three}(-3)^{ii}+2(-iii)+1 \\ &= \frac{i}{three}(9)-vi+i \\ &= 3-six+one \\ &= -2 \\ \therefore & (-three;-2) \terminate{align*}
parallel to the line \(y=4x-2\).
\begin{align*} \text{Gradient of tangent }&= f'(10) \\ f(x)&=(2x-1)^{two} \\ &= 4x^{ii}-4x+1 \\ \therefore f'(x)&= 8x-four \\ \text{Tangent is parallel to } y&=4x-ii \\ \therefore one thousand&=4 \\ \therefore f'(x) = 8x-4 &= four \\ 8x &= 8 \\ x & = 1\\ \text{For } 10=i: \quad y & = (two(1)-1)^{2} \\ & = ane \end{align*}
Therefore, the tangent is parallel to the given line at the bespeak \((1;1)\).
perpendicular to the line \(2y+x-4=0\).
\begin{align*} \text{Perpendicular to } 2y + x - 4 &= 0 \\ y&= -\frac{1}{2}x+2\\ \therefore \text{ slope of } \perp \text{ line } & = two \quad (m_1 \times m_2 = -one) \\ \therefore f'(x) &= 8x-4 \\ \therefore 8x-iv &=2\\ 8x&=half dozen\\ 10&=\frac{3}{four} \\ \therefore y&=\left[two\left(\frac{3}{4}\right)-1\right]^{2} \\ &=\frac{1}{4} \\ \therefore \left(\frac{3}{4};\frac{1}{4}\correct) \end{align*}
Therefore, the tangent is perpendicular to the given line at the betoken \(\left(\frac{3}{4};\frac{one}{4}\right)\).
Depict a graph of \(f\), indicating all intercepts and turning points.
Complete the square:
\brainstorm{align*} y&=-[x^{2}-4x+3] \\ &=-[(x-2)^{2}-4+3] \\ &=-(x-2)^{2}+1\\ \text{Turning point}:&(two;1) \cease{align*} \(\text{Intercepts:}\\ y_{\text{int}}: x = 0, y = -3 \\ x_{\text{int}}: y=0, \\ -x^{2} +4x -iii = 0 \\ ten^{ii} - 4x + 3 = 0 \\ (ten-3)(x-1) = 0 \\ x=iii \text{ or } x=1 \\ \text{Shape: "pout" } (a < 0) \\\) 
Detect the equations of the tangents to \(f\) at:
- the \(y\)-intercept of \(f\).
- the turning point of \(f\).
- the point where \(x = \text{4,25}\).
- \begin{align*} y_{\text{int}}: (0;-3) \\ m_{\text{tangent}} = f'(x) &= -2x + 4 \\ f'(0) &=-two(0) + 4 \\ \therefore m &=4\\ \text{Tangent }y&=4x+c\\ \text{Through }(0;-3) \therefore y&=4x-3 \stop{align*}
- \brainstorm{align*} \text{Turning point: } (ii;1) \\ m_{\text{tangent}} = f'(2) &= -2(2) + 4 \\ &=0\\ \text{Tangent equation } y &= 1 \end{marshal*}
- \begin{align*} \text{If } x &=\text{4,25} \\ f(\text{4,25})&=-\text{4,25}^{2}+4(\text{4,25})-iii \\ &= -\text{four,0625} \\ m_{\text{tangent}} \text{ at } ten&= \text{4,25} \\ grand&=-ii(\text{4,25})+4\\ &=-\text{4,5} \\ \text{Tangent }y&=-\text{4,v}x+c\\ \text{Through }(\text{4,25};-\text{iv,0625}) \\ -\text{4,0625}&=-\text{4,5}(\text{iv,25})+c\\ \therefore c&= \text{fifteen,0625} \\ y&=-\text{4,five}ten+\text{15,0625} \stop{marshal*}
Draw the three tangents above on your graph of \(f\).
Write downwards all observations about the 3 tangents to \(f\).
Tangent at \(y_{\text{int}}\) (blue line): gradient is positive, the office is increasing at this betoken.
Tangent at turning point (light-green line): gradient is zippo, tangent is a horizontal line, parallel to \(x\)-axis.
Tangent at \(x=\text{4,25}\) (purple line): slope is negative, the function is decreasing at this point.
Source: https://www.siyavula.com/read/maths/grade-12/differential-calculus/06-differential-calculus-04