MATH SOLVE

4 months ago

Q:
# Please help with calculus

Accepted Solution

A:

For each of these questions, you need to find the derivative [tex]y'[/tex] or [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex]. The slope of the tangent to these curves at the point [tex](a,b)[/tex] is the value of [tex]y'[/tex] when [tex]x=a[/tex] and [tex]y=b[/tex]. It's also important to know that if the slope of a line is [tex]m\neq0[/tex], then the slope of any line normal/perpendicular to this line is [tex]-\dfrac1m[/tex].###[tex]y=\dfrac{6x+3}{3x^2+6x+4}[/tex]The derivative is[tex]y'=\dfrac{(3x^2+6x+4)(6x+3)'-(6x+3)(3x^2+6x+4)'}{(3x^2+6x+4)^2}=\dfrac{6(3x^2+6x+4)-(6x+3)(6x+6)}{(3x^2+6x+4)^2}[/tex][tex]y'=\dfrac{6-18x-18x^2}{(3x^2+6x+4)^2}[/tex]When [tex]x=1[/tex], we get a slope of[tex]y'=\dfrac{6-18-18}{(3+6+4)^2}=-\dfrac{30}{169}[/tex]###[tex]y=x^2+9x+16[/tex]The derivative is[tex]y'=2x+9[/tex]and so the tangent line at (1, 9) has slope[tex]y'=2+9=11[/tex]The line normal to this has slope [tex]-\dfrac1{11}[/tex]. The point-slope and slope-intercept forms of this line are[tex]y-9=-\dfrac1{11}(x-1)\implies y=-\dfrac x{11}+\dfrac{100}{11}[/tex]###[tex]y=9x-14[/tex]The derivative is[tex]y'=9[/tex]so the slope of any line tangent to the curve is 9. The line that passes through (3, 4) is[tex]y-4=9(x-3)\impleis y=9x-23[/tex]