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Kendall tried to find all the equations of vertical lines tangent to the curve given by 
x^(2)+2xy^(2)=25. This is her solution:
Step 1: Finding an expression for 
(dy)/(dx).

(dy)/(dx)=(-x-y^(2))/(2xy)
Step 2: Forming a system of equations.

{[x^(2)+2xy^(2)=25],[2xy=0],[-x-y^(2)!=0]:}
Step 3: Solving the system.

x=-5,x=0, and 
x=5
Is Kendall's solution correct? If not, at which step did she make a mistake?
Choose 1 answer:
(A) The solution is correct.
(B) Step 1 is incorrect.
(C) Step 2 is incorrect.
(D) Step 3 is incorrect.

Kendall tried to find all the equations of vertical lines tangent to the curve given by x2+2xy2=25 x^{2}+2 x y^{2}=25 . This is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=xy22xy \frac{d y}{d x}=\frac{-x-y^{2}}{2 x y} \newlineStep 22: Forming a system of equations.\newline{x2+2xy2=252xy=0xy20 \left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlinex=5,x=0 x=-5, x=0 , and x=5 x=5 \newlineIs Kendall's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.

Full solution

Q. Kendall tried to find all the equations of vertical lines tangent to the curve given by x2+2xy2=25 x^{2}+2 x y^{2}=25 . This is her solution:\newlineStep 11: Finding an expression for dydx \frac{d y}{d x} .\newlinedydx=xy22xy \frac{d y}{d x}=\frac{-x-y^{2}}{2 x y} \newlineStep 22: Forming a system of equations.\newline{x2+2xy2=252xy=0xy20 \left\{\begin{array}{l} x^{2}+2 x y^{2}=25 \\ 2 x y=0 \\ -x-y^{2} \neq 0 \end{array}\right. \newlineStep 33: Solving the system.\newlinex=5,x=0 x=-5, x=0 , and x=5 x=5 \newlineIs Kendall's solution correct? If not, at which step did she make a mistake?\newlineChoose 11 answer:\newline(A) The solution is correct.\newline(B) Step 11 is incorrect.\newline(C) Step 22 is incorrect.\newline(D) Step 33 is incorrect.
  1. Forming Equations: Forming a system of equations.\newlineKendall's system of equations is based on the conditions for a vertical tangent line. For a vertical line, the slope is undefined, which means dydx\frac{dy}{dx} should be undefined. This occurs when the denominator of our derivative is zero, so 2xy=02xy = 0. Additionally, the numerator must not be zero to avoid the indeterminate form 00\frac{0}{0}, so xy20-x - y^2 \neq 0.\newlineKendall's system of equations is:\newline11. x2+2xy2=25x^2 + 2xy^2 = 25 (original equation)\newline22. 2xy=02xy = 0 (condition for vertical tangent)\newline33. xy20-x - y^2 \neq 0 (to avoid indeterminate form)\newlineKendall's system of equations is correct.
  2. Solving the System: Solving the system.\newlineKendall's solution to the system is x=5x = -5, x=0x = 0, and x=5x = 5.\newlineLet's check these solutions:\newlineFrom the second equation, 2xy=02xy = 0, we can see that either x=0x = 0 or y=0y = 0 for the equation to hold true. However, if y=0y = 0, then the first equation becomes x2=25x^2 = 25, which gives us x=±5x = \pm5. If x=0x = 0, then the first equation becomes x=0x = 000, which is not true. Therefore, x=0x = 011 cannot be x=0x = 022.\newlineKendall's inclusion of x=0x = 0 as a solution is incorrect. The correct solutions should only be x=5x = -5 and x=5x = 5, as these are the x-values where the curve x=0x = 066 has vertical tangents.\newlineKendall made a mistake in this step.

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