smc-1182-10-Find gradient

Calculus, SPEC2 2022 VCAA 10 MC

Consider the curve given by `5 x^2 y-3 x y+y^2=10`.

The equation of the tangent to this curve at the point `(1, m)`, where `m` is a real constant, will have a negative gradient when

`m \in R \backslash[-1,0]`
`m=-\sqrt{11}-1 \ text {only}`
`m \in R \backslash(-1,0]`
`m=\sqrt{11}-1 \ text[only]`
`m=-\sqrt{11}-1 or m=\sqrt{11}-1`

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`E`

Show Worked Solution

`5 x^2 y-3 x y+y^2=10`

`text{At}\ (1,m):\ \ 5m-3m+m^2=10`

`text{Solve (by CAS):}\ \ m=-1+-sqrt{11}`

`text{Check the value of the derivative function at both values of}\ m\ (x=1)`

`text{By CAS, both values (gradients) are negative.}`

`=>E`

Calculus, SPEC1 2022 VCAA 7

A curve has equation `x cos(x+y)=(pi)/(48)`.

Find the gradient of the curve at the point `((pi)/(24),(7pi)/(24))`. Give your answer in the form `(asqrtb-pi)/(pi)`, where `a,b in Z`. (3 marks)

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Show Answers Only

`(8sqrt3-pi)/pi`

Show Worked Solution

`text{Using implicit differentiation and the chain and product rules:}`

`d/(dx)(cos(x + y))`

`text{Let}\ \ u = x + y\ \ \=>\ \ y = cos(u)`

`(du)/(dx) = 1 +dy/dx,\ \ (dy)/(du) = -sin(u)`

`cos (x+y)-x\ sin (x+y)(dy/dx+1)`	`=0`
`cos (x+y)-x\ sin(x+y)(dy/dx)-x\ sin(x+y)`	`=0`

`x\ sin(x+y)(dy/dx)`	`=cos (x+y)-x\ sin(x+y)`
`dy/dx`	`=(cos (x+y))/(x\ sin(x+y)) – (x\ sin(x+y))/(x\ sin(x+y))`
	`=(cos (x+y))/(x\ sin(x+y))-1`

`text{At}\ ((pi)/(24),(7pi)/(24))\ \ \=>\ \ x+y = (pi)/(24)+(7pi)/(24) = (pi)/(3)`

`dy/dx`	`=cos(pi/3)÷[(pi)/(24)sin(pi/3)]-1`
	`=(1/2)/((pi)/(24) xx sqrt3/2)-1`
	`=24/(sqrt3pi)-1`
	`=(24-sqrt3pi)/(sqrt3pi) xx (sqrt3/sqrt3)`
	`=(24sqrt3-3pi)/(3pi)`
	`=(8sqrt3-pi)/pi`

♦ Mean mark 50%.

Calculus, SPEC2 2023 VCAA 1

Viewed from above, a scenic walking track from point \(O\) to point \(D\) is shown below. Its shape is given by

\(f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}\)

The minimum turning point of section \(O A B C\) occurs at point \(A\). Point \(B\) is a point of inflection and the curves meet at point \(C(1,0)\). Distances are measured in kilometres.

Show that \(a=-1\) and \(b=0\). (1 mark)

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Verify that the two curves meet smoothly at point \(C\). (2 marks)

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i. Find the coordinates of point \(A\). (1 mark)

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ii. Find the coordinates of point \(B\). (1 mark)

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The return track from point \(D\) to point \(O\) follows an elliptical path given by

\(x=2 \cos (t)+2, y=(e-2) \sin (t)\), where \(t \in\left[\dfrac{\pi}{2}, \pi\right]\).

Find the Cartesian equation of the elliptical path. (2 marks)

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Sketch the elliptical path from \(D\) to \(O\) on the diagram above. (1 mark)

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i. Write down a definite integral in terms of \(t\) that gives the length of the elliptical path from \(D\) to \(O\). (1 mark)

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ii. Find the length of the elliptical path from \(D\) to \(O\).
Give your answer in kilometres correct to three decimal places. (1 mark)

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a. \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

\(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)

b. \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)

c.i. \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)

c.ii. \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)

d. \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)

f.i. \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)

f.ii. \(\text{Length}\ = 2.255\)

Show Worked Solution

a. \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

\(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)

b. \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)

c.i. \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} \)

\(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)

c.ii. \(\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} \)

\(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)

d. \(\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) \)

\(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)

f.i. \(\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) \)

\(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)

f.ii. \(\text{Evaluate the integral in part f.i.}\)

\(\Rightarrow \text{Length}\ = 2.255\)

Calculus, SPEC2 2023 VCAA 9 MC

The position of a particle moving in the Cartesian plane, at time \(t\), is given by the parametric equations

\(x(t)=\dfrac{6 t}{t+1}\) and \(y(t)=\dfrac{-8}{t^2+4}\), where \(t \geq 0\).

What is the slope of the tangent to the path of the particle when \(t=2\) ?

\(-\dfrac{1}{3}\)
\(-\dfrac{1}{4}\)
\(\dfrac{1}{3}\)
\(\dfrac{3}{4}\)
\(\dfrac{4}{3}\)

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\(D\)

Show Worked Solution

\(\text{At}\ \ t=2\ \text{(by calc):}\)

\(\dfrac{dx}{dt}=\dfrac{2}{3}, \ \dfrac{dy}{dt}=\dfrac{1}{2} \)

\(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} \)

\(\Rightarrow D\)

Calculus, SPEC1 2021 VCAA 5

Find the gradient of the curve with equation `e^x e^(2y) = 2e^4` at the point `(2, 1)`. (3 marks)

Show Answers Only

`-1/10`

Show Worked Solution

`e^x e^(2y) + e^(4y^2) = 2e^4`

`e^x · 2e^(2y) · (dy)/(dx) + e^x · e^(2y) + e^(4y^2) · 8y · (dy)/(dx)`	`= 0`
`(dy)/(dx)(e^x ·2e^(2y) + e^(4y^2) · 8y)`	`= -e^x · e^(2y)`

`(dy)/(dx) = (-e^x · e^(2y))/(e^x · 2e^(2y) + e^(4y^2) ·8y)`

`text(At)\ \ (2, 1):`

`(dy)/(dx)`	`= (-e^2 · e^2)/(e^2 · 2e^2 + e^4 · 8)`
	`= (-e^4)/(e^4(2 + 8))`
	`= -1/10`

Calculus, SPEC1 2011 VCAA 10

Consider the relation `y log_e (x) = e^(2y) + 3x - 4.`

Evaluate `(dy)/(dx)` at the point `(1, 0).` (4 marks)

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`-3/2`

Show Worked Solution

`y log_e x = e^(2y) + 3x – 4`

`text(Using implicit differentiation:)`

`d/(dx)(y ln(x))`	`= d/(dx)(e^(2y)) + d/(dx)(3x) – d/(dx)(4)`
`dy/dx*ln(x) + y(1/x)`	`= 2e^(2y)*dy/dx + 3`

`text(At)\ \ (1,0):`

`dy/dx xx ln1 + 0`	`= 2 e^0 * dy/dx+ 3`
`2*dy/dx`	`= -3`
`:. dy/dx`	`=- 3/2`

Calculus, SPEC1 2011 VCAA 5

For the curve with parametric equations

`x = 4 sin (t) - 1`

`y = 2 cos (t) + 3`

Find `(dy)/(dx)` at the point `(1, sqrt 3 + 3).` (3 marks)

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`– sqrt 3/6`

Show Worked Solution

`(dx)/(dt) = 4cos(t)`

`(dy)/(dt) = −2sin(t)`

`(dy)/(dx)`	`= ((dy)/(dt))/((dx)/(dt))`
	`= (-sin t)/(2cos t)`

`x = 4 sin (t) – 1\ \ text{(given)}\ \ =>\ \ sin(t)=(x+1)/4`

`y = 2 cos (t) + 3\ \ text{(given)}\ \ =>\ \ 2cos(t)=(y-3)`

`=> dy/dx= (-(x + 1)/4)/(y – 3)`

`text(At)\ \ (1, sqrt 3 + 3):`

`:. (dy)/(dx)`	`= (-1/2)/(sqrt3+3 -3)`
	`= -1/(2sqrt3)`
	`= – sqrt3/6`

Calculus, SPEC1 2012 VCAA 6

Find the gradient of the tangent to the curve `xy^2 + y + (log_e (x - 2))^2 = 14` at the point `(3, 2).` (3 marks)

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`(dy)/(dx) = -4/13`

Show Worked Solution

`xy^2 + y + (log_e (x – 2))^2 = 14`

`text(Using implicit differentiation:)`

`d/(dx) (xy^2) + d/(dx) (y) + d/(dx) ((ln (x – 2))^2) = d/(dx)(14)`

`d/(dx) (x) ⋅ y^2 + d/(dx) (y^2) ⋅ x + (dy)/(dx) + 2 xx 1/(x – 2) xx ln(x – 2) = 0`

`y^2 + 2xy* (dy)/(dx) + (dy)/(dx) + (2 ln (x – 2))/(x – 2) = 0`

`text{At (3,2):}`

`2^2 + 2(2)(3) m + m + (2 ln(1))/1`	`= 0`
`4 + 12m + m`	`= 0`
`13m`	`= -4`
`:. m`	`= -4/13`

Graphs, SPEC1 2013 VCAA 4

State the maximal domain and the range of `y = arccos(1 - 2x).` (2 marks)
Sketch the graph of `y = arccos(1 - 2x)` over its maximal domain. Label the endpoints with their coordinates. (2 marks)
Find the gradient of the tangent to the graph of `y = arccos (1 – 2x)` at `x = 1/4.` (2 marks)

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`text(Maximal domain)\ \ [0, 1];\ \ \ text(Range)\ \ [0, pi]`
`4/sqrt3`

Show Worked Solution

a. `text(Maximal domain:)`

`1 – 2x`	`∈ [−1, 1]`
`-2x`	`∈ [−2, 0]`
`x`	`∈ [0, 1]`

`text(When)\ \ x = 0,\ \ y = cos^-1 1= 0;`

`text(When)\ \ x = 1,\ \ y = cos^-1 (-1)= pi`

`:.\ text(Range is)\ \ [0, pi]`

c. `y = cos^-1 (1 – 2x),`

`(dy)/(dx)`	`= (-1(-2))/sqrt(1 – (1 – 2x)^2)`
	`= 2/sqrt(1 – (1 – 2x)^2)`

`text(At)\ \ x = 1/4,`

`m_T`	`= 2/sqrt(1 – (1 – 1/2)^2)`
	`=2/sqrt(3/4)`
	`= 4/sqrt 3`
	`= (4 sqrt 3)/3`

Calculus, SPEC1 2013 VCAA 6

Find the value of `c`, where `c in R`, such that the curve defined by

`y^2 + (3e^{(x - 1)})/(x - 2) = c`

has a gradient of 2 where `x = 1.` (4 marks)

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`-3/4`

Show Worked Solution

`text(Using implicit differentiation:)`

`d/(dx)(y^2) + d/(dx)((3e^(x – 1))/(x – 2))`	`= d/(dx)(c)`
`2y(dy)/(dx) + ((x – 2)3e^(x – 1) – 3e^(x – 1))/(x – 2)^2`	`= 0`
`2y(dy)/(dx) +(3e^(x – 1) (x – 3))/(2y(x – 2)^2)`	`=0`

`:. (dy)/(dx) = -(3e^(x – 1) (x – 3))/(2y(x – 2)^2)`

`text(When)\ \ x=1,\ \ dy/dx=2`

`2=(-6)/(-2y)\ \ =>\ \ y=3/2`

`text(Substituting into curve equation:)`

`(3/2)^2 + (3e^0)/(1 – 2)`	`= c`
`9/4 – 3`	`= c`
`c`	`= -3/4`

Calculus, SPEC1-NHT 2017 VCAA 3

Find the gradient of the curve with equation `x = sin (y/15)` when `x = 1/4`. Give your answer in the form `a sqrt b`, where `a, b \ in Z^+`. (3 marks)

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`m = 4 sqrt 15`

Show Worked Solution

`x`	`= sin (y/15)`
`sin^(-1)x`	`= y/15`
`y`	`= 15 sin^(-1)x`
`dy/dx`	`= 15/(sqrt(1-x^2))`

`text(When)\ \ x=1/4:`

`dy/dx`	`= 15/(sqrt(1-(1/4)^2))`
	`= 15/(sqrt(15/16))`
	`= 60/sqrt15`
	`=4sqrt15`

`:. m = 4 sqrt 15`

Calculus, SPEC1 2015 VCAA 9

Consider the curve represented by `x^2 - xy + 3/2 y^2 = 9.`

Find the gradient of the curve at any point `(x, y).` (2 marks)
Find the equation of the tangent to the curve at the point `(3, 0)` and find the equation of the tangent to the curve at the point `(0, sqrt 6).`

Write each equation in the form `y = ax + b.` (2 marks)
Find the acute angle between the tangent to the curve at the point `(3, 0)` and the tangent to the curve at the point `(0, sqrt 6).`

Give your answer in the form `k pi`, where `k` is a real constant (2 marks)

Show Answers Only

`(dy)/(dx) = (2x – y)/(x – 3y)`
`y = 2(x – 3);\ \ \ y = 1/3 x + sqrt 6`
`theta = pi/4`

Show Worked Solution

a.	`d/(dx)(x^2) – d/(dx)(xy) + 3/2* d/(dx) (y^2)`	`= 0`
	`2x – x(dy)/(dx) – y + 3/2(2y)(dy)/(dx)`	`= 0`
	`(dy)/(dx)(−x + 3y)`	`= y – 2x`
	`:. (dy)/(dx)`	`= (y – 2x)/(3y – x)`

b. `m_{(3,0)} = (0 – 6)/(0 – 3) = 2`

`:.\ text{Equation of tangent at (3, 0):}`

`y = 2(x – 3)`

`= 2x – 6`

`m_{(0,sqrt6)} = (sqrt6 – 0)/(3sqrt6 – 0) = 1/3`

`:.\ text{Equation of tangent at}\ (0,sqrt6):`

`y -sqrt6`	`= 1/3(x – 0)`
`y`	`=1/3 x + sqrt6`

c. `m_1 = 2 = tan(theta_1), \ \ m_2 = 1/3 = tan(theta_2)`

`alpha`	`= theta_1 – theta_2`
	`= tan^(−1)(2) – tan^(−1)(1/3)`

`tan(alpha)`	`= tan(tan^(−1)(2) – tan^(−1)(1/3))`
	`= (tan(tan^(−1)(2)) – tan(tan^(−1)(1/3)))/(1 + tan(tan^(−1)(2)tan(tan^(−1)(1/3))))`
	`= (2 – 1/3)/(1 + 2/3)`
	`= 1`

`:. alpha = pi/4\ \ \ (alpha ∈ (0, pi/2))`

Calculus, SPEC1-NHT 2018 VCAA 9

i. Given that `cot(2 theta) = a`, show that `tan^2(theta) + 2a tan(theta) - 1 = 0`. (2 marks)
ii. Show that `tan(theta) = -a +- sqrt(a^2 + 1)`. (1 mark)
iii. Hence, show that `tan(pi/12) = 2 - sqrt 3`, given that `cot(2 theta) = sqrt 3`, where `theta in (0, pi)`. (1 mark)
Find the gradient of the tangent to the curve `y = tan (theta)` at `theta = pi/12`. (2 marks)
A solid of revolution is formed by rotating the region between the graph of `y = tan(theta)`, the horizontal axis, and the lines `theta = pi/12` and `theta = pi/3` about the horizontal axis.

Find the volume of the solid of revolution. (3 marks)

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1. `text(Proof)\ \ text{(See Worked Solutions)}`
2. `text(Proof)\ \ text{(See Worked Solutions)}`
3. `text(Proof)\ \ text{(See Worked Solutions)}`
`8 – 4 sqrt 3`
`pi (2 sqrt 3 – 2 – pi/4)`

Show Worked Solution

a.i.	`1/(tan 2 theta)`	`= a`
	`1`	`= a tan (2 theta)`
	`1`	`= a ((2 tan (theta))/(1 – tan^2(theta)))`

	`1/a (1 – tan^2 (theta))`	`= 2 tan (theta)`
	`1 – tan^2 (theta)`	`= 2a tan (theta)`
	`:. tan^2 (theta) + 2 a tan (theta) – 1`	`=0\ \ text(… as required)`

a.ii.	`[tan^2 (theta) + 2a tan (theta) + a^2] – a^2 – 1`	`= 0`
	`(tan (theta) + a)^2`	`= a^2 + 1`
	`tan (theta) + a`	`= +- sqrt(a^2 + 1)`
	`:. tan (theta)`	`= -a +- sqrt(a^2 + 1)`

a.iii. `theta in (0, pi) \ => \ 2 theta in (0, 2 pi)`

`text(S)text(ince)\ \ cot(2theta)=sqrt3\ \ \ =>\ \ \ tan(2theta)=1/sqrt3`

	`:. 2 theta`	`= pi/6, pi + pi/6`
	`theta`	`= pi/12, (7 pi)/12`
	`tan (theta)`	`= – sqrt 3 +- sqrt((sqrt 3)^2 + 1)`
		`= – sqrt 3 +- sqrt(3 + 1)`
		`= – sqrt 3 +- 2`

`:. tan (pi/12) = 2 – sqrt 3,\ \ \ \ (tan (pi/12) > 0)`

b.	`y`	`=tan(theta)`
	`y prime`	`= sec^2 (theta)`
		`= 1 + tan^2 (theta)`

`y prime (pi/12)`	`= 1 + tan^2 (pi/12)`
	`= 1 + (2 – sqrt 3)^2`
	`= 1 + 4 – 4 sqrt 3 + 3`
	`= 8 – 4 sqrt 3`

c.	`V`	`= pi int_(pi/12)^(pi/3) y^2\ d theta`
		`= pi int_(pi/12)^(pi/3) tan^2 (theta)\ d theta`
		`= pi int_(pi/12)^(pi/3) (1 + tan^2 (theta) – 1)\ d theta`
		`= pi int_(pi/12)^(pi/3) (sec^2 (theta) – 1)\ d theta`
		`= pi [tan (theta) – theta]_(pi/12)^(pi/3)`
		`= pi (tan (pi/3) – pi/3 – (tan (pi/12) – pi/12))`
		`= pi (sqrt 3 – (4 pi)/12 – (2 – sqrt 3) + pi/12)`
		`= pi (2 sqrt 3 – 2 – pi/4)\ \ text(u³)`

Calculus, SPEC1 2018 VCAA 3

Find the gradient of the curve with equation `2x^2 sin(y) + xy = pi^2/18` at the point `(pi/6, pi/6)`.

Give your answer in the form `a/(pi sqrt b + c)`, where `a, b` and `c` are integers. (4 marks)

Show Answers Only

`m = (-18)/(pi sqrt 3 + 6)`

Show Worked Solution

`d/(dx) (2x^2 sin (y)) + d/(dx) (xy) = d/(dx) (pi^2/18)`

`4x sin(y) + 2x^2 cos(y) (dy)/(dx) + y + x (dy)/(dx)`	`=0`
`(4 pi)/6 sin (pi/6) + (2 pi^2)/36 cos (pi/6) m + pi/6 + pi/6 m`	`=0`
`(2 pi)/3 xx 1/2 + pi^2/18 xx sqrt 3/2 m + pi/6 + pi/6 m`	`=0`
`pi/3 + pi/6 + ((pi^2 sqrt 3)/36 + pi/6) m`	`=0`
`pi/2 + pi/36 (pi sqrt 3 + 6) m`	`=0`

`(pi m)/36 (pi sqrt 3 + 6)`	`= (-pi)/2`
`m (pi sqrt 3 + 6)`	`= -18`
`:. m`	`= (-18)/(pi sqrt 3 + 6)`