The equation of a curve from its gradient questions with answers

Questions

1. Question 1Polynomial gradients

   The gradient of a curve is given by $\dfrac{dy}{dx} = 6x^{2} - 4x + 1$. The curve passes through the point $(1, 4)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Integrate dy/dx to get y (add 1 to each power, divide by the new power) and add c$y = \dfrac{6x^{3}}{3} - \dfrac{4x^{2}}{2} + x + c$$= 2x^{3} - 2x^{2} + x + c$
   2. Step 2: Use the point the curve passes through: put its x and y in to find c$4 = 2(1)^{3} - 2(1)^{2} + 1 + c$$4 = 1 + c$$c = 3$
   3. Step 3: Write the equation with c put in$y = 2x^{3} - 2x^{2} + x + 3$
2. Question 2Polynomial gradients

   The gradient of a curve is given by $\dfrac{dy}{dx} = 9x^{2} + 2x - 3$. The curve passes through the point $(2, 3)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Integrate dy/dx to get y (add 1 to each power, divide by the new power) and add c$y = \dfrac{9x^{3}}{3} + \dfrac{2x^{2}}{2} - 3x + c$$= 3x^{3} + x^{2} - 3x + c$
   2. Step 2: Use the point the curve passes through: put its x and y in to find c$3 = 3(2)^{3} + (2)^{2} - 3(2) + c$$3 = 22 + c$$c = -19$
   3. Step 3: Write the equation with c put in$y = 3x^{3} + x^{2} - 3x - 19$
3. Question 3Negative and fractional powers

   The gradient of a curve is given by $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{x}} + 3x^{2}$. The curve passes through the point $(1, -3)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Write every term as a power of x$\dfrac{dy}{dx} = x^{-\frac{1}{2}} + 3x^{2}$
   2. Step 2: Integrate dy/dx to get y (add 1 to each power, divide by the new power) and add c$y = \dfrac{x^{\frac{1}{2}}}{\dfrac{1}{2}} + \dfrac{3x^{3}}{3} + c$$= 2x^{\frac{1}{2}} + x^{3} + c$
   3. Step 3: Write it with positive indices$y = 2\sqrt{x} + x^{3} + c$
   4. Step 4: Use the point the curve passes through: put its x and y in to find c$-3 = 2\sqrt{(1)} + (1)^{3} + c$$-3 = 3 + c$$c = -6$
   5. Step 5: Write the equation with c put in$y = 2\sqrt{x} + x^{3} - 6$
4. Question 4Negative and fractional powers

   The gradient of a curve is given by $\dfrac{dy}{dx} = 2x - \dfrac{4}{x^{2}}$. The curve passes through the point $(1, -3)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Write every term as a power of x$\dfrac{dy}{dx} = 2x - 4x^{-2}$
   2. Step 2: Integrate dy/dx to get y (add 1 to each power, divide by the new power) and add c$y = \dfrac{2x^{2}}{2} - \dfrac{4x^{-1}}{-1} + c$
   3. Step 3: Write it with positive indices$y = x^{2} + \dfrac{4}{x} + c$
   4. Step 4: Use the point the curve passes through: put its x and y in to find c$-3 = (1)^{2} + \dfrac{4}{(1)} + c$$-3 = 5 + c$$c = -8$
   5. Step 5: Write the equation with c put in$y = x^{2} + \dfrac{4}{x} - 8$
5. Question 5Exponential and trigonometric gradients

   The gradient of a curve is given by $\dfrac{dy}{dx} = 6e^{-3x}$. The curve passes through the point $(0, 1)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Integrate: eᵇˣ integrates to eᵇˣ ÷ b, and a number k to kx. Add c$y = -2e^{-3x} + c$
   2. Step 2: Use the point the curve passes through: put its x and y in to find c$1 = -2e^{-3(0)} + c$$1 = -2 + c$$c = 3$
   3. Step 3: Write the equation with c put in$y = -2e^{-3x} + 3$
6. Question 6Exponential and trigonometric gradients

   The gradient of a curve is given by $\dfrac{dy}{dx} = 6\cos 3x + 2$, where $x$ is in radians. The curve passes through the point $(0, 5)$. Find the equation of the curve.

   Show the worked solution

   1. Step 1: Integrate: cos bx integrates to (sin bx) ÷ b, and a number k to kx. Add c$y = 2\sin 3x + 2x + c$
   2. Step 2: Use the point the curve passes through: put its x and y in to find c$5 = 2\sin 3(0) + 2(0) + c$$5 = 0 + c$$c = 5$
   3. Step 3: Write the equation with c put in$y = 2\sin 3x + 2x + 5$