Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
Simplify each side of the following equation before applying the addition propertyx − 9 + 6 = −5 − 4
