Respuesta :
Given the coordinates of two points P and Q, we can calculate the distance between them using the formula:
[tex]\begin{gathered} \begin{cases}P={(x_P},y_P) \\ Q={(x_Q},y_Q)\end{cases} \\ . \\ d=\sqrt{(x_P-x_Q)^2+(y_P-y_Q)^2} \end{gathered}[/tex]In this case, we want the smallest distance between a point in the curve and the point (0, 4)
Then, we know that there is a point that we can call Q = (x, y) that is the closest to the point (0, 4). We can write, using the distance formula:
[tex]d=\sqrt{(x-0)^2+(y-4)^2}=\sqrt{x^2+(y-4)^2}[/tex]The equation given is:
[tex]y=5x+1[/tex]We want to rewrite the distance formula to include the equation of the curve. Since there is a term 'x²', we can solve the equation for x and square on both sides:
[tex]\begin{gathered} y=5x+1 \\ . \\ y-1=5x \\ , \\ x=\frac{y}{5}-\frac{1}{5} \\ . \\ x^2=(\frac{y}{5}-\frac{1}{5})^2 \end{gathered}[/tex]Now we can substitute in the distance equation:
[tex]d=\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}[/tex]We can see that this is a distance function for any point of the curve to the point (0, 4). This is actually a function of y.
[tex]d(y)=\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}[/tex]Now, we can apply calculus to find the minimum of this function. Let's take the first derivative:
[tex]d^{\prime}(y)=\frac{\frac{2}{5}(\frac{y}{5}-\frac{1}{5})+2(y-4)}{2\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}}[/tex]Simplify:
[tex]d^{\prime}^(y)=\frac{26y-101}{25\sqrt{(\frac{y}{5}-\frac{1}{5})^2+(y-4)^2}}[/tex]And since we want to find a minimum, we need to also calculate the second derivative:
[tex]d^{\prime}^{\prime}(y)=\frac{26\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}-\frac{(2(y-4)+\frac{2}{5}(\frac{y}{5}-\frac{1}{5})(26y-101)}{2\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}}}{25((y-4)^2+(\frac{y}{5}-\frac{1}{5})^2)}[/tex]Simplify:
[tex]d^{\prime}^{\prime}(y)=\frac{9}{25((y-4)^2+(\frac{y}{5}-\frac{1}{5})^2)^{\frac{3}{2}}}[/tex]Now, we need to find the critical points of the function. The critical points are the x values where the first derivative is 0.
Then:
[tex]d^{\prime}(y)=\frac{26y-101}{25\sqrt{(y-4)^2+(\frac{y}{5}-\frac{1}{5})^2}}[/tex]Is a quotient, For a quotient to be 0, the only way this is possible is for the numerator to be 0. Then:
[tex]\begin{gathered} 26y-101=0 \\ . \\ y=\frac{101}{26} \end{gathered}[/tex]And now, to see if this critical point is a minimum, we evaluate it in the second derivative, if the second derivative is positive in this critical point, the function has a minimum at that point:
[tex]d^{\prime}^{\prime}(\frac{10}{26})=\frac{9}{25((\frac{101}{26}-4)^2+(\frac{101}{26}-\frac{1}{5})^2)^{\frac{3}{2}}}\approx1.76746[/tex]Then, the function d(y) has a minimum at y = 101/26
Now, we need to find the x coordinate of this point. We use the equation of the curve:
[tex]\begin{gathered} \frac{101}{26}=5x+1 \\ . \\ x=(\frac{101}{26}+1)\cdot\frac{1}{5}=\frac{15}{26} \end{gathered}[/tex]Thus, the answer to the point in the curve that is the closest to (0, 4) is:
[tex](\frac{15}{26},\frac{101}{26})[/tex]