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