Let S be the focus of y^2 = 4x and a point P | vedclass

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(A) The focus $S$ of the parabola $y^2 = 4x$ is $(1, 0)$. Let $P$ be a point on the parabola,so $P = (T^2, 2T)$.The vector $\vec{SP} = (T^2 - 1)\hat{i

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$\sqrt{2}$

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(A) The focus $S$ of the parabola $y^2 = 4x$ is $(1, 0)$. Let $P$ be a point on the parabola,so $P = (T^2, 2T)$.The vector $\vec{SP} = (T^2 - 1)\hat{i} + 2T\hat{j}$.The line is $x + y - 1 = 0$. The normal vector to the line is $\vec{n} = \hat{i} + \hat{j}$. The unit normal vector is $\hat{u} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.The projection of $\vec{SP}$ on the line $x + y = 1$ is the component of $\vec{SP}$ along the direction parallel to the line. The direction vector of the line is $\vec{v} = \hat{i} - \hat{j}$.The projection $L$ is given by $\frac{\vec{SP} \cdot \vec{v}}{|\vec{v}|} = \frac{(T^2 - 1) - 2T}{\sqrt{2}} = \frac{T^2 - 2T - 1}{\sqrt{2}}$.Differentiating with respect to $t$: $\frac{dL}{dt} = \frac{1}{\sqrt{2}} (2T - 2) \frac{dT}{dt}$.Given $\frac{dx}{dt} = 4$. Since $x = T^2$,$\frac{dx}{dt} = 2T \frac{dT}{dt} = 4$. At $P(4, 4)$,$T = 2$,so $4 \frac{dT}{dt} = 4 \implies \frac{dT}{dt} = 1$.Substituting $\frac{dT}{dt} = 1$ and $T = 2$ into the derivative of $L$:$\frac{dL}{dt} = \frac{1}{\sqrt{2}} (2(2) - 2)(1) = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Let $S$ be the focus of $y^2 = 4x$ and a point $P$ is moving on the curve such that its abscissa is increasing at the rate of $4 \text{ units/sec}$. Then the rate of increase of the projection of $SP$ on the line $x + y = 1$ when $P$ is at $(4, 4)$ is:

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