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Calculus, 2ADV C3 2021 HSC 13

Find the exact gradient of the tangent to the curve  `y = x tan x`  at the point where  `x = pi/3`.  (3 marks)

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`sqrt3 + (4pi)/3`

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`y = x tan x`

`(dy)/(dx) = tan x + x sec^2 x`

`text(Find)\ \ m\ \ text(when)\ \ x = pi/3:`

`(dy)/(dx)` `= tan\ pi/3 + pi/3 · 1/(cos^2\ pi/3)`
  `= sqrt3 + pi/3 · 1/(1/4)`
  `= sqrt3 + (4pi)/3`

Calculus, 2ADV C3 2021 HSC 10 MC

The line  `y = mx`  is a tangent to the curve  `y = cos x`  at the point where  `x = a`, as shown in the diagram.
 

Which of the following statements is true?

  1. `m < 1/a < 1/(2pi)`
  2. `1/(2pi) < m < 1/a`
  3. `1/(2pi) < 1/a < m`
  4. `m < 1/(2pi) < 1/a`
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`B`

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`text(By Elimination:)`

♦♦♦ Mean mark 27%.

`a < 2pi \ => \ 1/a > 1/(2pi)`

`->\ text(Eliminate A)`
 

`m ~~ 1/(2pi) ~~ 1/6`

`1/a\ text{is much closer to 1 (by inspection)}`

`1/a > m`

`->\ text(Eliminate C)`
 

`text(At)\ \ x = 2pi,`

`y = cos 2pi = 1\ \ text(and)\ \ y = mx > 1`

`m > 1/(2pi)`

`:. 1/(2pi) < m < 1/a`

 
`=>B`

Calculus, 2ADV C3 2018 HSC 12b

Find the equation of the tangent to the curve  `y = cos 2x`  at  `x = pi/6`.  (3 marks)

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`y = -sqrt 3 x + (sqrt 3 pi + 3)/6`

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`y` `= cos 2x`
`(dy)/(dx)` `= -2 sin 2x`

 
`text(When)\ \ x = pi/6:`

`y` `= cos  pi/3 = 1/2`
`(dy)/(dx)` `= -sin  pi/3 = -sqrt 3`

 
`text(Equation of tangent)\ \ m = -sqrt 3,\ text(through)\ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= -sqrt 3 (x – pi/6)`
`y` `= -sqrt 3 x + (sqrt 3 pi)/6 + 1/2`
`:. y` `= -sqrt 3 x + (sqrt 3 pi + 3)/6`

Calculus, 2ADV C3 2016 HSC 11f

Find the gradient of the tangent to the curve  `y = tan x`  at the point where  `x = pi/8.`

Give your answer correct to 3 significant figures.  (2 marks)

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`1.17`

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`y = tan x = (sin x)/(cos x)`

`(dy)/(dx) = sec^2 x`

`text(When)\ \ x = pi/8,`

`:. (dy)/(dx)` `= sec^2(pi/8)`
  `= 1/(cos^2(pi/8))`
  `= 1.1715…`
  `= 1.17\ \ text{(to 3 sig fig)}`

Calculus, 2ADV C3 2007 HSC 2c

The point  `P (pi, 0)`  lies on the curve  `y = x sinx`. Find the equation of the tangent to the curve at  `P`.  (3 marks)

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`y = – pi x + pi^2`

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`y = x sin x`

`(dy)/(dx)` `= x xx d/(dx) (sin x) + d/(dx) x xx sin x`
  `= x  cos x + sin x`

 
`text(When)\ \ x = pi`

`(dy)/(dx)` `= pi xx cos pi + sin pi`
  `= pi (-1) + 0`
  `= – pi`

 
`text(Equation of line,)\ \ m = – pi,\ text(through)\ P(pi, 0):`

`y – y_1` `= m(x – x_1)`
`y – 0` `= – pi(x – pi)`
`:. y` `= – pi x + pi^2`

Calculus, 2ADV C3 2006 HSC 2c

Find the equation of the tangent to the curve  `y = cos 2x`  at the point whose `x`-coordinate is  `pi/6`.  (3 marks)

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`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

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`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6`

`y` `= cos (2 xx pi/6)`
  `= cos (pi/3)`
  `= 1/2`

 

`dy / dx` `= -2 sin (pi/3)`
  `= -2 xx sqrt 3 / 2`
  `= – sqrt 3`

 
`text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= – sqrt 3 ( x – pi/6)`
`y – 1/2` `= – sqrt 3 x + (sqrt 3 pi)/6`
`y` `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`