A curve is defined by the parametric equations - AQA - A-Level Maths Mechanics - Question 3 - 2017 - Paper 2
Question 3
A curve is defined by the parametric equations.
$$x = t^3 + 2,
y = t^2 - 1$$
3 (a) Find the gradient of the curve at the point where $t = -2$.
3 (b) Find a Carte... show full transcript
Worked Solution & Example Answer:A curve is defined by the parametric equations - AQA - A-Level Maths Mechanics - Question 3 - 2017 - Paper 2
Step 1
Find the gradient of the curve at the point where $t = -2$
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Answer
To find the gradient of the curve defined by the parametric equations, we need to compute
dxdy=dtdxdtdy.\n
Calculate (\frac{dx}{dt}:)
Given (x = t^3 + 2), we differentiate:
dtdx=3t2.
Calculate (\frac{dy}{dt}:)
Given (y = t^2 - 1), we differentiate:
dtdy=2t.
Substitute (t = -2) into both derivatives:
For (\frac{dx}{dt}:)
dtdx=3(−2)2=3⋅4=12.
For (\frac{dy}{dt}:)
dtdy=2(−2)=−4.
Now, calculate the gradient:
dxdy=dtdxdtdy=12−4=−31.
Thus, the gradient of the curve at the point where t=−2 is (-\frac{1}{3}).
Step 2
Find a Cartesian equation of the curve
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Answer
To eliminate the parameter (t), we express (y) in terms of (x):
From the equation for (x):
x=t3+2⇒t3=x−2.
Now substituting (t) into the equation for (y):
From (y = t^2 - 1), we need (t):
t=3x−2.
Substitute into (y):
y=(3x−2)2−1=−(x−2)2/31.
Thus, the Cartesian equation of the curve is:
y=(x−2)2/3−1.
