Problem on finding a tangent vector to a curve

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Let $y = \frac{1}{2} x^2+\frac{1}{2}$. For what value of $c$ is $\mathbf{i} + c \mathbf{j}$ a tangent vector to $y(x)$ at $x=1$?

  • Solution

    The vector is tangent to the curve if it has the same 'slope' in the $xy$ plane as the tangent line to $y(x)$ at $x=1$.
    Recall that
    The derivative $\frac{dy}{dx}(x)$ gives the slope of the tangent line to $y(x)$ at $x$.
    Computing that $\frac{dy}{dx} = x$, the slope of $y(x)$ at $x=1$ is $1$.
    The 'slope' of the vector is given by its rise of $c$ over its run of $1$:
    Hence, $\frac{c}{1}=1$, and the tangent vector is $\mathbf{i} + \mathbf{j}.$

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