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  • . The slope of tangent to the curve x equals t2 plus 3t minus 8, y equals 2t2 minus 2t minus 5 at the point (2, minus1) is:

The slope of tangent to the curve x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 at the point (2, -1) is:

A. \frac{22}{7}

B. \frac{6}{7}

C. -\frac{6}{7}

D. -6

Answers (1)

Curve of the given equation is  x = t^2 + 3t - 8, y = 2t^2 - 2t - 5

With respect to t, while differentiating on both sides, we get

\frac{d(x)}{d t}=\frac{d\left(t^{2}+3 t-8\right)}{d t}$
After application of the sum rule of differentiation, we get
\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}+\frac{\mathrm{d}(3 \mathrm{t})}{\mathrm{dt}}+\frac{\mathrm{d}(-8)}{\mathrm{dt}}$

Constant's derivative is 0, so above equation becomes
\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}+3 \frac{\mathrm{d}(\mathrm{t})}{\mathrm{dt}}+0$

Power Rule application leads to
\\ \frac{d(x)}{d t}=2 t+3 \ldots \ldots$(i) \\$y=2 t^{2}-2 t-5$
With respect to t, we differentiate on both side and get

\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{\mathrm{d}\left(2 \mathrm{t}^{2}-2 \mathrm{t}-5\right)}{\mathrm{dt}}$

Sum Rule application leads to
\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{\mathrm{d}\left(2 \mathrm{t}^{2}\right)}{\mathrm{dt}}-\frac{\mathrm{d}(2 \mathrm{t})}{\mathrm{dt}}+\frac{\mathrm{d}(-5)}{\mathrm{dt}}$
The Constant's derivative is 0, so the equation becomes
\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=2 \frac{\mathrm{d}\left(\mathrm{t}^{2}\right)}{\mathrm{dt}}-2 \frac{\mathrm{d}(\mathrm{t})}{\mathrm{dt}}+0$
Applying power rule
\frac{d(y)}{d t}=4 t-2 \ldots$(ii)

We know,
\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Substitute values from equation (i) and (ii)
\frac{d y}{d x}=\frac{4 t-2}{2 t+3}$

The point through which the curve passes is (2,-1), now, substitute the same and get

\\ {x = t\textsuperscript{2} + 3t - 8}\\ { \Rightarrow 2= t\textsuperscript{2} + 3t - 8}\\ { \Rightarrow t\textsuperscript{2} + 3t - 8-2=0}\\ { \Rightarrow t\textsuperscript{2} + 3t - 10=0}\\

Split the middle term

\\ { \Rightarrow t\textsuperscript{2} + 5t-2t - 10=0}\\ { \Rightarrow t(t+ 5) -2(t+5)=0}\\ { \Rightarrow (t+ 5) (t-2)=0}\\ { \Rightarrow t+5=0 or t-2=0}\\ { \Rightarrow t=-5or t=2 \ldots \ldots \ldots .(iii)}\\ {y = 2t\textsuperscript{2} - 2t - 5}\\ { \Rightarrow -1=2t\textsuperscript{2} - 2t - 5}\\ { \Rightarrow 2t\textsuperscript{2} - 2t - 5+1=0}\\ { \Rightarrow 2t\textsuperscript{2} - 2t - 4=0}\\

Take 2 as common

\Rightarrow t\textsuperscript{2} - t - 2=0

Split the middle term again

\\ { \Rightarrow t\textsuperscript{2} - 2t +t- 2=0}\\ { \Rightarrow t(t- 2)+1(t- 2)=0}\\ { \Rightarrow (t- 2) (t+1) = 0}\\ { \Rightarrow (t- 2) =0 or (t+1) = 0}\\ { \Rightarrow t=2 or t=-1 \ldots \ldots .. (iv)}\\

In equation (iii) and (iv), 2 is common

So, t=2

So, the slope of the tangent at t=2 is as follows

\\ \left(\frac{d y}{d x}\right)_{t=2}=\frac{4 t-2}{2 t+3}$ \\$\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{t}=2}=\frac{4(2)-2}{2(2)+3}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{8-2}{4+3}$ \\$\Rightarrow\left(\frac{d y}{d x}\right)_{t=2}=\frac{6}{7}$
Therefore, the slope of tangent at the point (2,-1) is \frac{6}{7}$
So, the correct answer is option B

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