Find an equation for the line tangent to the curve at the point defined by the given value of t.
-$$\mathrm { x } = \mathrm { t } , \mathrm { y } = \sqrt { 2 \mathrm { t } } , \mathrm { t } = 18$$
A) $y = \frac { 1 } { 6 } x$
B) $y = - \frac { 1 } { 6 } x - 3$
C) $y = \frac { 1 } { 6 } x + 3$
D) $y = - \frac { 1 } { 6 } x$

Question

Quizplus · Accepted Answer

The derivative of the curve is $ \frac { dy } { dx } = \frac { 1 } { \sqrt { 2 t } } $. At $ t = 18 $, the slope of the tangent line is $ \frac { 1 } { \sqrt { 2 ( 18 ) } } = \frac { 1 } { 6 } $. The point of tangency is $ ( 18 , 3 ) $. So the equation of the tangent line is $ y - 3 = \frac { 1 } { 6 } ( x - 18 ) $, which simplifies to $ y = \frac { 1 } { 6 } x + 3 $.

Find an Equation for the Line Tangent to the Curve x=t,y=2t,t=18\mathrm { x } = \mathrm { t } , \mathrm { y } = \sqrt { 2 \mathrm { t } } , \mathrm { t } = 18x=t,y=2t​,t=18

Find an Equation for the Line Tangent to the Curve $\mathrm { x } = \mathrm { t } , \mathrm { y } = \sqrt { 2 \mathrm { t } } , \mathrm { t } = 18$