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