The slope of the tangent to the curve x=t2+3t−8,y=2t2−2t−

Discussion Forum : Application Of Derivatives

Question -

The slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at the point t = 2 is

Options:

A .

\frac{7}{6}

B .

\frac{5}{6}

C .

\frac{6}{7}

D .

1

Answer: Option C
:
C
We have,

$\frac{d x}{d t} = 2 t + 3 a n d \frac{d y}{d t} = 4 t - 2 \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 t - 2}{2 t + 3}$
Thus, slope of the tangent to the curve at the point t = 2 is
${[\frac{d y}{d x}]}_{t - 2} = \frac{4 (2) - 2}{2 (2) + 3} = \frac{6}{7}$
Thus, slope of the tangent to the curve at the point t = 2 is
Hence (c) is the correct answer

Was this answer helpful ?

Next Question

Discussion Forum : Application Of Derivatives

Submit Your Solution hear: