Answer to Question #110933 in Calculus for please1.3

Question

[img]https://upload.cc/i1/2020/04/20/RAD7Mo.jpg[/img]


question is in picture,R=4

Answer 1

a) $f(x)=x^2(3x^3-2)^{1/3}; P=(1,-2)$

Slope $= dy/dx=2x(3x^3-2)^{1/3}+x^2(3x^2)/(3x^3-2)^{2/3}$

$=2x(3x^3-2)^{1/3}+3x^4/(3x^3-2)^{2/3}$

$=2*1(3*1^3-2)^{1/3}+3*1^4/(3*1^3-2)^{2/3}$ at the point (1,-2)

$=2+3=5$

Thus, the tangent has a slope of 5 and passes through the point (1,-2)

$y+2=5(x-1) \implies y=5x-7$ ---(Answer)

b) $f(x)=sin^3xcos^4x; P=(0,R)$

Slope $=dy/dx=3sin^2xcos^4x(cosx)+4sin^3xcos^3x(-sinx)$

$=3sin^2xcos^5x-4sin^4xcos^3x$

$=sin^2xcos^3x(3cos^2x-4sin^2x)$

$=sin^20cos^30(3cos^20-4sin^20)$ at the point (0,R)

$=0$

Thus, the tangent has a slope of 0 and passes through the point (0,R).

$y-R=0*(x-0) \implies y=R$

Given : $R=4 \implies y=4$ ---(Answer)