Calculus
Finding a curve from its gradient
In this note
- 14.10Find the equation of a curve from \(\frac{dy}{dx}\) and a point on it.
- 14.10Find a curve from \(\frac{d^{2}y}{dx^{2}}\) (two constants) or from the gradient of the normal.
- 14.6Then find and classify the curve's stationary points.
Before you start: Integration basics, e and trig, stationary points. Next: Areas under curves.
Formula bank
From dy/dx and a point
Integrating \(\frac{dy}{dx}\) gives a whole family of curves (one for each c). The point on the curve picks out the one you want.
General methodHow to find a curve from its gradient
- Integrate dy/dx, with + c.
- Put the point's x and y into this equation.
- Solve for c and write the full equation y = … .
Example \(\frac{dy}{dx}=6x^{2}-4x+1\) and the curve passes through \((1,5)\). Find its equation.
Integrate, then use the point.
- \(y=2x^{3}-2x^{2}+x+c\)
- \(5=2-2+1+c\Rightarrow c=4\)
Answer \(y=2x^{3}-2x^{2}+x+4\)
Example \(\frac{dy}{dx}=\sqrt{2x+1}\); the curve passes through \((4,10)\).
\(\int(2x+1)^{1/2}dx=\tfrac13(2x+1)^{3/2}\).
- \(10=\tfrac13(9)^{3/2}+c=9+c\)
Answer \(y=\tfrac13(2x+1)^{3/2}+1\)
Try it yourself
Show answer
Answer \(y=2\sin2x+3\)
Hint
Show answer
Answer \(y=2\ln x+3\)
From the second derivative
Given \(\frac{d^{2}y}{dx^{2}}\), integrate TWICE. Each integration brings its own constant, so you need two facts: a gradient at a point (for the first constant) and a point on the curve (for the second).
General methodHow to find a curve from d²y/dx²
- Integrate once: dy/dx = … + c₁.
- Use the GRADIENT you are given to find c₁.
- Integrate again: y = … + c₂.
- Use the POINT to find c₂.
Example \(\frac{d^{2}y}{dx^{2}}=6x-2\). The curve has gradient 3 at the point \((1,2)\). Find its equation.
Two constants, two facts.
- \(\frac{dy}{dx}=3x^{2}-2x+c_1\); \(3=3-2+c_1\Rightarrow c_1=2\)
- \(y=x^{3}-x^{2}+2x+c_2\); \(2=1-1+2+c_2\Rightarrow c_2=0\)
Answer \(y=x^{3}-x^{2}+2x\)
Then the stationary points
Example \(\frac{dy}{dx}=3x^{2}-12\); the curve passes through \((0,5)\). Find its stationary points and their nature.
Curve first, then dy/dx = 0.
- \(y=x^{3}-12x+5\)
- \(3x^{2}-12=0\Rightarrow x=\pm2\)
- \(\frac{d^{2}y}{dx^{2}}=6x\): at \(x=2\) it's 12 > 0 (min); at \(x=-2\) it's −12 < 0 (max)
Answer minimum \((2,-11)\), maximum \((-2,21)\)
Ways they test it
Type 1 Curve from dy/dx
"A curve is such that \(\frac{dy}{dx}=\ldots\). It passes through (…). Find its equation." (4-5 marks)
Spot it a gradient function + one point.
- Integrate + c.
- Point → c.
Type 2 Then a stationary point
"…Find the coordinates of the stationary point on the curve." (6 marks)
Spot it a follow-on part.
- dy/dx = 0 from the GIVEN derivative.
- y from your curve.
For example \(\frac{dy}{dx}=(5x-2)^{1/3}\)
- \(x=\tfrac25\)
Answer \(\big(\tfrac25,4\big)\)
Type 3 From d²y/dx²
"\(\frac{d^{2}y}{dx^{2}}=5\cos2x\). The gradient is \(\tfrac34\) at (…). Find the curve." (8 marks)
Spot it second derivative + a gradient + a point.
- Integrate, gradient → c₁.
- Integrate, point → c₂.
Type 4 From the normal
"The gradient of the normal at (x, y) is …" (5 marks)
Spot it 'normal'.
- \(-1\div\) it to get dy/dx.
- Then as normal.
From the past papers
Past paper 1
How to start: Integrate the bracket to a power (divide by the new power AND by 5), find c from the point, then dy/dx = 0.
✓ Show the solution have a go on paper first
- \(y=\dfrac{(5x-2)^{4/3}}{\tfrac43\times5}+c=\tfrac{3}{20}(5x-2)^{4/3}+c\).
- At \((2,\tfrac{32}{5})\): \(\tfrac{32}{5}=\tfrac{3}{20}(8)^{4/3}+c=\tfrac{3}{20}(16)+c\Rightarrow c=4\).
- Stationary: \((5x-2)^{1/3}=0\Rightarrow x=\tfrac25\); then \(y=0+4=4\).
- Answer: \(\big(\tfrac25,4\big)\).
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(\frac{dy}{dx}=0\Rightarrow x=\tfrac25\) | B1 | must use the first derivative | |
| \(\tfrac{3}{20}(5x-2)^{4/3}\) | B2 | M1 for \(k(5x-2)^{4/3}\) | |
| use of the point to find c | M1 | ||
| \(c=4\) | A1 | ||
| \(\big(\tfrac25,4\big)\) | B1 | ||
| Total: 6 marks | |||
Examiner tip Paper 1 has no calculator: \(8^{4/3}=(\sqrt[3]8)^{4}=16\).
Past paper 2
How to start: Two integrations. The gradient \(\tfrac34\) fixes the first constant, the point fixes the second. Radians.
✓ Show the solution have a go on paper first
- \(\frac{dy}{dx}=\tfrac52\sin2x+c_1\); at \(x=-\tfrac{\pi}{12}\): \(\tfrac34=\tfrac52\sin\big(-\tfrac{\pi}{6}\big)+c_1=-\tfrac54+c_1\Rightarrow c_1=2\).
- \(y=-\tfrac54\cos2x+2x+c_2\).
- At \(\big(-\tfrac{\pi}{12},\tfrac{5\pi}{4}\big)\): \(\tfrac{5\pi}{4}=-\tfrac54\cdot\tfrac{\sqrt3}{2}-\tfrac{\pi}{6}+c_2\Rightarrow c_2=\tfrac{17\pi}{12}+\tfrac{5\sqrt3}{8}\).
- Answer: \(y=-\tfrac54\cos2x+2x+\tfrac{17\pi}{12}+\tfrac{5\sqrt3}{8}\) (\(c_2\approx5.53\)).
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(\frac{dy}{dx}=\tfrac52\sin2x+c\) | M1A1 | M1 for \(k\sin2x\) | |
| use of the gradient | M1 | ||
| \(c=2\) | A1 | ||
| \(y=-\tfrac54\cos2x+2x+d\) | M1A1 | ||
| use of the point | M1 | ||
| \(d=\tfrac{17\pi}{12}+\tfrac{5\sqrt3}{8}\) (or 5.53) | A1 | must have the full equation | |
| Total: 8 marks | |||
Examiner tip The commonest slip: using the point's y-value to find the FIRST constant. The first constant always comes from a gradient.
Mistakes to avoid
- ✗ Leaving out + c, so the point can't be used✓ + c every time you integrate
- ✗ Using the point to find the constant in dy/dx✓ gradient → c₁, point → c₂
- ✗ Normal gradient used as the curve's gradient✓ \(m=-1\div m_{\text{normal}}\)
- ✗ Stationary point from y = 0✓ from dy/dx = 0
Summary
- Integrate dy/dx with + c; the point gives c.
- d²y/dx²: integrate twice; gradient → c₁, point → c₂.
- Normal gradient m: the curve's gradient is \(-1/m\).
If you see …
- 'passes through'find c.
- 'gradient … at the point'find the constant in dy/dx first.
Practice
Practice: easy, medium, hard
Easy
Show answer
Answer \(y=2x^{2}-3x+1\)
Show answer
Answer \(y=\tfrac32e^{2x}+\tfrac52\)
Show answer
Answer \(y=2x^{3}-2x^{2}+x+4\)
Medium
Guide me
What is it asking? a curve
What do we know? \(\tfrac13(2x+1)^{3/2}+c\)
Plan x = 4 gives 9 + c
Show working
- \(c=1\)
Answer \(y=\tfrac13(2x+1)^{3/2}+1\)
Guide me
What is it asking? a curve
What do we know? \(\frac{dy}{dx}=x^{3}-4x+c_1\), \(c_1=0\)
Plan \(y=\tfrac14x^{4}-2x^{2}+c_2\)
Show working
- \(c_2=5\)
Answer \(y=\tfrac14x^{4}-2x^{2}+5\)
Guide me
What is it asking? a value
What do we know? \(y=x^{3}-12x+5\)
Plan x = 2 is the minimum
Show working
- (2, −11)
Answer −11
Hard
Guide me
What is it asking? a curve
What do we know? \(c_1=2\)
Plan \(c_2=0\)
Show working
- \(y=x^{3}-x^{2}+2x\)
Answer \(y=x^{3}-x^{2}+2x\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(3x^{2}-2x+c\) | M1 | ||
| \(c=2\) | A1 | ||
| \(x^{3}-x^{2}+2x+d\) | M1 | ||
| \(d=0\) | A1 | ||
| Total: 4 marks | |||
Guide me
What is it asking? a curve
What do we know? \(\frac{dy}{dx}=-\tfrac13\cos3x+\tfrac13\)
Plan \(y=-\tfrac19\sin3x+\tfrac13x+1\)
Show working
- \(c_1=\tfrac13\), \(c_2=1\)
Answer \(y=-\tfrac19\sin3x+\tfrac13x+1\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(-\tfrac13\cos3x+c\) | M1A1 | ||
| \(c=\tfrac13\) | A1 | ||
| second integration and d = 1 | M1A1 | ||
| Total: 5 marks | |||
Guide me
What is it asking? a value
What do we know? \(y=\tfrac{3}{20}(5x-2)^{4/3}+c\)
Plan \(c=4\); stationary at \(x=\tfrac25\)
Show working
- \(\big(\tfrac25,4\big)\)
Answer 4
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| integral | B2 | ||
| c = 4 | M1A1 | ||
| (2/5, 4) | B1 | ||
| Total: 5 marks | |||
Can I do this?
- I can find a curve from dy/dx and a point (or the normal's gradient).
- I can find a curve from d²y/dx² using a gradient and a point.
- I can then find and classify stationary points.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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