Lesson G5
G5 Implicit and parametric differentiation Quiz: AQA Maths, Unit 7
20 questions
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Lesson G5, Implicit and parametric differentiation: 20 multiple choice questions for the AQA Maths (7357), Unit 7: Differentiation, written with Revision Ninja.
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The 20 questions
-
For the implicit curve x^2 + y^2 = 25, what is dy/dx?
- -y/x
- 2x
- x/y
- -x/y
-
Differentiating y^3 with respect to x, what is the result?
- 3y dy/dx
- 3y^2
- y^3 dx/dy
- 3y^2 dy/dx
-
For x = t^2 and y = t^3, what is dy/dx?
- 3t/2
- 3/2
- 2t/3
- 3/(2t^2)
-
The formula dy/dx = (dy/dt)/(dx/dt) for parametric curves comes from which rule?
- The chain rule
- Integration by parts
- The product rule
- The quotient rule applied to x only
-
Differentiate implicitly the relation xy = 4.
- dy/dx = 4
- y dy/dx = 4
- x dy/dx = 4
- y + x dy/dx = 0
-
Why is implicit differentiation needed for a curve such as x^2 + y^2 = 1?
- Because the curve is a straight line and needs no differentiation
- Because the curve has no gradient at any point
- Because y cannot easily be written as a single function of x, so the chain rule is applied to both sides
- Because y is constant on the curve
-
A curve is given by x = 2t and y = t^2. What is dy/dx?
- 1/t
- t
- 2t
- t^2
-
Find dy/dx for x^2 + y^2 = 25 at the point (3, 4).
- -3/4
- -4/3
- 3/4
- 4/3
-
Find the gradient of x^3 + y^3 = 9 at the point (2, 1).
- -4
- 4
- -1/4
- -2
-
For x = 3 cos t and y = 3 sin t, what is dy/dx at t = pi/4?
- -sqrt(2)
- 1
- -1
- 0
-
For x = t + 1 and y = t^2 - 4t, what is dy/dx at t = 2?
- 0
- -2
- 2
- 1
-
Find dy/dx for x^2 y + y^2 = 6x.
- (6 - 2xy)/(x^2 - 2y)
- (3 - xy)/(x^2 + y)
- (6 - 2xy)/(x^2 + 2y)
- (6 + 2xy)/(x^2 + 2y)
-
For the curve y^2 = 4x, what is dy/dx in terms of y?
- y/2
- 2/x
- 4/y
- 2/y
-
For x = e^t and y = t, what is dy/dx?
- t e^t
- e^t
- 1/t
- e^(-t)
-
For x = t^3 and y = t^2, what is the gradient at t = 1?
- 1
- 2/3
- 3/2
- 2
-
For the curve x^3 + y^3 = 6xy, what is dy/dx at the point (3, 3)?
- -1
- -3
- 3
- 1
-
A curve has x = t^2 + 1 and y = t^3 - t. What is the gradient at t = 1?
- 1/2
- 1
- 2
- -1
-
For x = sin t and y = cos 2t, what is dy/dx at t = pi/6?
- 2
- -sqrt(3)/2
- -1/2
- -2
-
Differentiate implicitly e^y = x + 1 to find dy/dx.
- x + 1
- 1/(x + 1)
- y/(x + 1)
- e^y
-
For x^2 + xy + y^2 = 7, what is dy/dx at the point (1, 2)?
- 4/5
- -4/5
- -5/4
- -1/2
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