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Exercise 7.2 · Q6

Q.Find the equations of the tangents to the curve y=1+x3y=1+x^3 for which the tangent is orthogonal with the line x+12y=12x+12y=12.

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"Orthogonal to the line" means the tangent's slope is the negative reciprocal of the line's slope; find that required slope first, then solve y′=(that value)y'=(\text{that value}).

Step 1. Slope of the line and the required (orthogonal) tangent slope.

x+12y=12⇒y=−x12+1x+12y=12\Rightarrow y=-\tfrac{x}{12}+1, slope =−112=-\tfrac{1}{12}. Orthogonal tangent slope =−1−1/12=12=-\dfrac{1}{-1/12}=12.

Step 2. Differentiate the curve and solve y′=12y'=12.

y=1+x3⇒y′=3x2=12⇒x2=4⇒x=±2y=1+x^3\Rightarrow y'=3x^2=12\Rightarrow x^2=4\Rightarrow x=\pm2. …

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