Answer to Question #247494 in Calculus for Lian

Question #247494
Find an equation of the line tangent to the curve y = 3x
2 − 1 and
perpendicular to the line x − 3y = 4.
1
Expert's answer
2021-10-06T17:27:27-0400

First we have to use the definition for the curve y1 to find m1:


"y_1=3x^2-1\n\\\\ y'_1=\\cfrac{dy_1}{dx}=6x=m_1"


"x-3y=4\n\\\\ 1-3y'_2=0 \\implies y'_2=m_2=1\/3\n\\\\ \\text{For perpendicular lines we have}\n\\\\ m_1m_2=-1\n\\\\ \\implies m_1=-1\/m_2=-1\/(1\/3)=-3"


After we find m1 = -3, we can use the definition for y'1 to find the point (x*,y*) to find the equation of the tangent line:


"-3=m_1=6x \\implies x^*=-1\/2; y^*_1=3(1\/2)^2-1=-1\/4"


At last, we substitute to find the equation of the tangent line:


"y-y_1^*=m_1(x-x^*)\n\\\\ y-(-1\/4)=(-3)(x-(-1\/2))\n\\\\ y+(1\/4)=-3x-(3\/2)\n\\\\ y=-3x-\\cfrac{7}{4} \\implies 4y+12x+7=0"


In conclusion, 4y + 12x + 7 = 0 is the equation of the tangent line to the function y = 3x2 - 1 at point P ( -1/2, -1/4 ).

Reference:

  • Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.

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