Question

Найдите угловой коэффициент угловой касательной к кривой y= -1/2x^2 в точке с ординатой y= -1.5

Answer 1

Что такое угловой коэффициент касательной ?  Это значение производной данной функции в точке касания.
y = -1/2x²
y' = 2/x³
y'(-1,5) = 2/(-1,5)³ = -16/27

Answer 2

Calculating the Slope of the Tangent Line

To find the slope of the tangent line to the curve \( y = -\frac{1}{2}x^2 \) at the point with the ordinate \( y = -1.5 \), we can use the derivative of the function \( y = -\frac{1}{2}x^2 \). The derivative will give us the slope of the tangent line at any given point on the curve.

The derivative of the function \( y = -\frac{1}{2}x^2 \) is given by:

\[ \frac{dy}{dx} = -x \]

Now, we can find the slope of the tangent line at the point with the ordinate \( y = -1.5 \) by substituting the value of \( x \) that corresponds to this ordinate into the derivative.

Calculating the Slope at the Given Point

Substituting the value of \( y = -1.5 \) into the derivative \( \frac{dy}{dx} = -x \), we get:

\[ -1.5 = -x \]

Solving for \( x \), we find:

\[ x = 1.5 \]

So, at the point where the ordinate is \( y = -1.5 \), the value of \( x \) is 1.5.

Finding the Slope of the Tangent Line

Now that we have the value of \( x \), we can find the slope of the tangent line by substituting this value into the derivative \( \frac{dy}{dx} = -x \).

\[ \frac{dy}{dx} \Bigg|_{x=1.5} = -1.5 \]

Therefore, the slope of the tangent line to the curve \( y = -\frac{1}{2}x^2 \) at the point with the ordinate \( y = -1.5 \) is -1.5.