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:


y1=3x21y1=dy1dx=6x=m1y_1=3x^2-1 \\ y'_1=\cfrac{dy_1}{dx}=6x=m_1


x3y=413y2=0    y2=m2=1/3For perpendicular lines we havem1m2=1    m1=1/m2=1/(1/3)=3x-3y=4 \\ 1-3y'_2=0 \implies y'_2=m_2=1/3 \\ \text{For perpendicular lines we have} \\ m_1m_2=-1 \\ \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=m1=6x    x=1/2;y1=3(1/2)21=1/4-3=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:


yy1=m1(xx)y(1/4)=(3)(x(1/2))y+(1/4)=3x(3/2)y=3x74    4y+12x+7=0y-y_1^*=m_1(x-x^*) \\ y-(-1/4)=(-3)(x-(-1/2)) \\ y+(1/4)=-3x-(3/2) \\ 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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