1. **State the problem:** We have the curve defined by the equation $$y=\sqrt{1-x^2}$$ and points A(0.6, 0.8), B(0.7, y_B), C(0.8, y_C), and D(0.9, y_D) on the curve. We need to verify if A lies on the curve, estimate the gradient at A, find gradients of line segments AD, AC, and AB, and compare these gradients. 2. **Verify point A lies on the curve:** Substitute $x=0.6$ into the curve equation: $$y=\sqrt{1-(0.6)^2}=\sqrt{1-0.36}=\sqrt{0.64}=0.8$$ Since the given y-coordinate of A is 0.8, point A lies on the curve. 3. **Find y-coordinates of B, C, and D:** - For B at $x=0.7$: $$y_B=\sqrt{1-(0.7)^2}=\sqrt{1-0.49}=\sqrt{0.51}\approx 0.714$$ - For C at $x=0.8$: $$y_C=\sqrt{1-(0.8)^2}=\sqrt{1-0.64}=\sqrt{0.36}=0.6$$ - For D at $x=0.9$: $$y_D=\sqrt{1-(0.9)^2}=\sqrt{1-0.81}=\sqrt{0.19}\approx 0.436$$ 4. **Estimate the gradient of the curve at A:** The gradient of the curve is the derivative of $y$ with respect to $x$: $$y=\sqrt{1-x^2}=(1-x^2)^{1/2}$$ Using the chain rule: $$\frac{dy}{dx}=\frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}$$ At $x=0.6$: $$\frac{dy}{dx} = \frac{-0.6}{\sqrt{1-0.36}}=\frac{-0.6}{0.8}=-0.75$$ So, the gradient of the curve at A is $-0.75$. 5. **Find gradients of line segments:** The gradient between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$m=\frac{y_2 - y_1}{x_2 - x_1}$$ - i) Gradient of AD: $$m_{AD}=\frac{y_D - y_A}{x_D - x_A}=\frac{0.436 - 0.8}{0.9 - 0.6}=\frac{-0.364}{0.3} \approx -1.213$$ - ii) Gradient of AC: $$m_{AC}=\frac{y_C - y_A}{x_C - x_A}=\frac{0.6 - 0.8}{0.8 - 0.6}=\frac{-0.2}{0.2}=-1$$ - iii) Gradient of AB: $$m_{AB}=\frac{y_B - y_A}{x_B - x_A}=\frac{0.714 - 0.8}{0.7 - 0.6}=\frac{-0.086}{0.1}=-0.86$$ 6. **Comment on the relationship:** The gradient of the curve at A is $-0.75$. The gradients of the line segments AB, AC, and AD are $-0.86$, $-1$, and $-1.213$ respectively. As the points get farther from A (from B to D), the gradient of the line segment deviates more from the curve's gradient at A. This shows that the tangent gradient at A is best approximated by the gradient of the line segment closest to A (AB), and the farther points give less accurate approximations. **Final answers:** - Point A lies on the curve. - Gradient of curve at A: $-0.75$ - Gradients of line segments: $m_{AB}=-0.86$, $m_{AC}=-1$, $m_{AD}=-1.213$ - The closer the point to A, the closer the line segment gradient to the curve gradient at A.