Question

The slope of the tangent to the curve y=root of 4-x^2 at the point where the ordinate and the abscissa are equal is

Answer 1

point where abscissa and coordinate are equal

${x}^{2} + {y}^{2} = 4 \\ y = x \\ {x}^{2} + {x}^{2} = 4 \\ x = y = \sqrt{2} \\ not \: - \sqrt{2} because \: negative \: numbers \: are \: not \: in \: domain \: of \: y \: \\ see \: the \: graph \\ (because \: y = \sqrt{4 - {x}^{2} } \: and \:result \: of \: square \: root \: is \: always \: positive \: hence \: y \: is \: positive)$



so the point is (√2,√2)

$slope \: \\ y = \sqrt{4 - {x}^{2} } \\ \frac{dy}{dx} = \frac{1}{2\sqrt{4 - {x}^{2} } } ( - 2x) \\ \frac{dy}{dx} = - \frac{x}{ \sqrt{4 - {x}^{2} } } \\ at \: x = \sqrt{2} \\ \\ slope = - \frac{ \sqrt{2} }{ \sqrt{2} } = 1$

mark as brainliest if helped

Answer 2

The slope of the tangent to the curve y=$\sqrt{4-x^2}$ at the point where the ordinate and the abscissa is equal is -1.

• Given curve is y = $\sqrt{4 - x²}$

Squaring both sides we get,

y² = 4 - x²

x² + y² = 4

• The given equation is a circle with radius 2.

• The center of the circle is (0,0) and radius is 2.

• Now to find the slope of the tangent to the curve , we find dy/dx.

Differentiating the equation with respect to x.

2x + 2y (dy/dx) = 0

dy/dx = -2x/2y = -x / y

• Now we have to find the slope where the abscissa and ordinate is equal.

That is x = y.

dy/dx = -1.

• Slope of the tangent to the curve where abscissa and ordinate is equal is -1.