Question
A curve is defined by the parametric equations
x = 3t − 2 sin t, y = 5t + 4 cos t,
where \(0\leqslant t\leqslant 2\pi \). At each of the points P and Q on the curve, the gradient of the curve is \(\frac{5}{2}\).
(a) Show that the values of t at P and Q satisfy the equation 10 cos t − 8 sin t = 5. [3]
(b) Express 10 cost − 8 sin t in the form \(R\cos (t+\alpha ), where \ R> 0> and \ 0< \alpha < \frac{1}{2}\pi \). Give the exact
value of R and the value of \(\alpha\) correct to 3 significant figures. [3]
(c) Hence find the values of t at the points P and Q. [4]
Answer/Explanation
Ans
(a) Carry out division at least as far as 3x2+ kx
Obtain quotient 3x2 – 4 – 4
Confirm remainder is 9 AG
(b) Integrate to obtain at least k1 x3 and k2 ln(3x + 2) terms
Obtain x3 – 2x2 – 4x + 3 ln(3x + 2)
(FT from quotient in part (a))
Apply limits correctly
Apply appropriate logarithm properties correctly
Obtain 125 ln64
(c) State or imply \(9x^{3}-6x^{2}-20x-8=(3x+2)(3x^{2}-4x-4) \)
(FT from quotient in part (a))
Attempt to solve cubic eqn to find positive value of x (or of 3ey )
Use logarithms to solve equation of form 3ey = k where k > 0 M1
Obtain \(\frac{1}{3}ln2\) or exact equivalent
Question
The diagram shows part of the curve with equation
\(y=4sin^2x+8sinx+3\).
where x is measured in radians. The curve crosses the x-axis at the point A and the shaded region is bounded by the curve and the lines x=0 and y=0.
(a) Find the exact x-coordinate of A.
(b) Find the exact gradient of the curve at A.
(c) Find the exact area of the shaded region.
Answer/Explanation
Ans:
(a) Solve equation y = 0 to find the value of x
Obtain \(\frac{7}{6}\pi\)
(b) Attempt first derivative using chain rule
Obtain \(\frac{dy}{dx}=8sinxcosx+8cos x\)
Substitute value from part (a) to find gradient \(-2\sqrt{3}\)
(c) Express integrand in the form \(k_1+k_2 cos 2x + k_3 sin x\)
Obtain correct 5 – 2cos 2x + 8sin x
Integrate to obtain 5x – sin2x – 8cos x
Apply limits 0 and their value from part (a) correctly
Obtain \(\frac{35}{6}\pi+\frac{7}{2}\sqrt{3}+8\) or exact equivalent
Question
A curve has equation
\(3x^2-y^2-4 In(2y+3)=26\).
Find the equation of the tangent to the curve at the point (3, -1).
Answer/Explanation
Ans:
Differentiate \(-y^2\) to obtain \(-2y\frac{dy}{dx}\)
Differentiate -4In(2y+3) to obtain \(\frac{-8}{2y+3}\frac{dy}{dx}\)
Attempt differentiation of all terms
Substitute x = 3, y =-1 to find numerical value of \(\frac{dy}{dx}\)
Obtain \(\frac{dy}{dx}=3\)
Obtain equation y = 3x – 10
Question
The diagram shows part of the curve with equation \(y=\frac{5x}{4x^3+1}\). The shaded region is bounded by the curve and the lines x=1, x=3 and y=0.
(a) Find \(\frac{dy}{dx}\) and hence find the the x-coordinate of the maximum point.
(b) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
(c) State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.
Answer/Explanation
Ans:
- Differentiate using quotient rule (or product rule)
Obtain \(\frac{5(4x^3+1)-60x^3}{(4x^3+1)^2}\)
Equate first derivative to zero and attempt solution
Obtain \(x=\frac{1}{2}\) - Use y values \(\frac{5}{5}, \frac{10}{33}, \frac{15}{109}\) or decimal equavalents
Use correct formula, or equivalent, with h=1
Obtain \(\frac{1}{2}(1+\frac{20}{30}+\frac{15}{109})\) or equivalent and hence 0.87 - State over-estimate with reference to top of each trapezium above curve