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Find the equation of the normal to the cury {:[(dy)/(dx)=1.84e^(0.5 x)" at "x=1],[(0.5 x)/(dx)(dy)/(d(1))=3.08]:}

249249. Find the equation of the normal to the cury\newlinedydx=1.84e0.5x at x=10.5xdxdyd(1)=3.08 \begin{array}{l} \frac{d y}{d x}=1.84 e^{0.5 x} \text { at } x=1 \\ \frac{0.5 x}{d x} \frac{d y}{d(1)}=3.08 \end{array}

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Q. 249249. Find the equation of the normal to the cury\newlinedydx=1.84e0.5x at x=10.5xdxdyd(1)=3.08 \begin{array}{l} \frac{d y}{d x}=1.84 e^{0.5 x} \text { at } x=1 \\ \frac{0.5 x}{d x} \frac{d y}{d(1)}=3.08 \end{array}
  1. Find Tangent Slope: First, we need to find the slope of the tangent to the curve at x=1x=1 using the given derivative.
  2. Calculate Tangent Slope: Substitute x=1x=1 into the derivative to find the slope of the tangent.\newline(dydx)=1.84e(0.5×1)(\frac{dy}{dx}) = 1.84e^{(0.5 \times 1)}\newline(dydx)=1.84e(0.5)(\frac{dy}{dx}) = 1.84e^{(0.5)}
  3. Calculate Exact Slope: Now, calculate the exact value of the slope at x=1x=1.(dydx)=1.84×e0.5(\frac{dy}{dx}) = 1.84 \times e^{0.5}
  4. Find Normal Slope: The slope of the normal is the negative reciprocal of the slope of the tangent. Slope of normal = −1/(1.84∗e0.5)-1 / (1.84 * e^{0.5})
  5. Find Equation of Normal: To find the equation of the normal, we need a point on the curve at x=1x=1. We don't have the original equation of the curve, so we can't find the yy-coordinate of the point. We need the yy-coordinate to proceed.

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