Tangents (Adv-X)

Calculus, 2ADV C1 2019 HSC 14d v1

The equation of the tangent to the curve `y = ae^(2x)+bx` at the point where `x = 0` is `y = 3x +2`.

Find the values of `a` and `b`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`b = -1,\ \ a = 2`

Show Worked Solution

`y `	`= ae^(2x)+bx`
`(dy)/(dx)`	`= 2ae^(2x)+b`

`text(When)\ \ x = 0,\ \ (dy)/(dx) = 3`

`2a + b`

`= 3\ …\ (1)`

`text(The point)\ (0, 2)\ text(lies on)\ y:`

`a(1) +b(0)`	`=2`
`a`	`= 2\ …\ (2)`

`text(Substitute into)\ (1)`

`2(2)+b`	`= 3`
`b`	`= -1`

Calculus, 2ADV C1 EO-Bank 2

Find the equations of the tangents to the curve `y = x^2-5x+6` at the points where the curve cuts the `x`-axis. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Where do the tangents intersect? (2 marks)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`y = −x+2`
`y = x-3`
`(5/2, −1/2)`

Show Worked Solution

a. `y`	`= x^2-5x+6`
	`= (x-2)(x-3)`

`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`

`(dy)/(dx) = 2x-5`

`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0`	`= -1(x-2)`
`y`	`= -x+2`

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y -0`	`= 1(x-3)`
`y`	`= x -3`

b. `text(Intersection occurs when:)`

`-x+2`	`= x-3`
`2x`	`= 5`
`x`	`= 5/2`

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 EO-Bank 1

Find the equation of the tangent to the curve \(y=e^{x^2+3x}\) at the point where \(x=1\). (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`y = 5e^4x-4e^4`

Show Worked Solution

\(y\)	\(=e^{x^2+3x}\)
`(dy)/(dx)`	\(=(2x+3)e^{x^2+3x}\)

`text(When)\ x = 1,\ \ (dy)/(dx) = 5e^4`

`text(Equation of tangent through)\ (1, e^4)`

`y-e^4`	`= 5e^4(x – 1)`
`y`	`= 5e^4x-4e^4`

Calculus, 2ADV C1 2023 HSC 14 v1

Find the equation of the tangent to the curve `y=x(3x+2)^2` at the point `(1,25)`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`y=55x-30`

Show Worked Solution

`y`	`=x(3x+2)^2`
`dy/dx`	`=6x(3x+2) + (3x+2)^2`
	`=(3x+2)(9x+2)`

`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1`	`=m(x-x_1)`
`y-25`	`=55(x-1)`
`y`	`=55x-30`

Calculus, 2ADV C1 2009 HSC 1d v1

Find the gradient of the tangent to the curve `y = 2x^3-5x^2 + 4` at the point `(2, 0)`. (2 marks)

Show Answers Only

`text(Gradient = 2.)`

Show Worked Solution

`y`	`= 2x^3-5x^2 + 4`
`dy/dx`	`= 6x^2-10x`

`text(At)\ x = 2:`

`dy/dx= 6(2)^2-10(2)=24-20=4`

`:.\ text(Gradient of tangent at)\ (2, 0) = 4.`

Calculus, 2ADV C1 EO-Bank 3 MC

At which point on the curve \(y = x^{2}-6x + 8\) can a normal be drawn such that it is inclined at 45\(^{\circ}\) to the positive \(x\)-axis?

\((1,3)\)
\((2,0)\)
\(\left(\dfrac{5}{2}, -\dfrac{3}{4}\right)\)
\((5,-7)\)

Show Answers Only

\(C\)

Show Worked Solution

\(y = x^{2}-6x + 8\)

\(y^{′} = 2x-6\)

If the normal is inclined at \(45^{\circ}\) to the positive x-axis then:

\(m_{\text{normal}} = \tan 45^{\circ} = 1\)

\(\text{Since } m_{\text{tangent}} \times m_{\text{normal}} = -1,\)

\(\therefore m_{\text{tangent}} = -1.\)

Find \(x\) when \(y^{′} = -1:\)

\(2x-6\)	\(=-1\)
\(2x\)	\(=5\)
\(x\)	\(=\dfrac{5}{2}\)

Find \(y:\)

\(y\)	\(= \left(\dfrac{5}{2}\right)^{2}-6\left(\dfrac{5}{2}\right) + 8\)
	\(=\dfrac{25}{4}-15 + 8\)
	\(= -\dfrac{3}{4}\)

\(\Rightarrow C\)