Tangent and Normal
Application of Derivatives

85393 The approximate value of \((1.002)^{\mathbf{3 0 0}}\) using differentiation is

1 1.8
2 1.6
3 1.2
4 1.4
Application of Derivatives

85394 Using differentiation, the approximate value of \(\sin 46^{\circ}\),
Given that \(1^{0}=0.0175^{\circ}\), is

1 0.07194
2 \(\frac{1.0175}{\sqrt{2}}\)
3 0.7194
4 \(\frac{0.0175}{\sqrt{2}}\)
Application of Derivatives

85395 The slope of normal to the curve \(x=\sqrt{t}\) and
\(\mathbf{y}=\mathbf{t}-\frac{1}{\sqrt{\mathbf{t}}} \text { at } \mathbf{t}=\mathbf{4} \text { is }\)
\(\begin{array}{llll}
\text { (a) } \frac{17}{4} \text { (b) } \frac{-17}{4} \text { (c) } \frac{4}{17} \text { (d) } \frac{-4}{17}
\end{array}\)

1 \(\frac{17}{4}\)
2 \(\frac{-17}{4}\)
3 \(\frac{4}{17}\)
4 \(\frac{-4}{17}\)
Application of Derivatives

85396 The equation of normal to the curve \(y=\log _{e} x\) at the point \(P(1,0)\) is

1 \(x+y=1\)
2 \(2 x+y=2\)
3 \(x-y=1\)
4 \(x-2 y=1\)
Application of Derivatives

85393 The approximate value of \((1.002)^{\mathbf{3 0 0}}\) using differentiation is

1 1.8
2 1.6
3 1.2
4 1.4
Application of Derivatives

85394 Using differentiation, the approximate value of \(\sin 46^{\circ}\),
Given that \(1^{0}=0.0175^{\circ}\), is

1 0.07194
2 \(\frac{1.0175}{\sqrt{2}}\)
3 0.7194
4 \(\frac{0.0175}{\sqrt{2}}\)
Application of Derivatives

85395 The slope of normal to the curve \(x=\sqrt{t}\) and
\(\mathbf{y}=\mathbf{t}-\frac{1}{\sqrt{\mathbf{t}}} \text { at } \mathbf{t}=\mathbf{4} \text { is }\)
\(\begin{array}{llll}
\text { (a) } \frac{17}{4} \text { (b) } \frac{-17}{4} \text { (c) } \frac{4}{17} \text { (d) } \frac{-4}{17}
\end{array}\)

1 \(\frac{17}{4}\)
2 \(\frac{-17}{4}\)
3 \(\frac{4}{17}\)
4 \(\frac{-4}{17}\)
Application of Derivatives

85396 The equation of normal to the curve \(y=\log _{e} x\) at the point \(P(1,0)\) is

1 \(x+y=1\)
2 \(2 x+y=2\)
3 \(x-y=1\)
4 \(x-2 y=1\)
Application of Derivatives

85393 The approximate value of \((1.002)^{\mathbf{3 0 0}}\) using differentiation is

1 1.8
2 1.6
3 1.2
4 1.4
Application of Derivatives

85394 Using differentiation, the approximate value of \(\sin 46^{\circ}\),
Given that \(1^{0}=0.0175^{\circ}\), is

1 0.07194
2 \(\frac{1.0175}{\sqrt{2}}\)
3 0.7194
4 \(\frac{0.0175}{\sqrt{2}}\)
Application of Derivatives

85395 The slope of normal to the curve \(x=\sqrt{t}\) and
\(\mathbf{y}=\mathbf{t}-\frac{1}{\sqrt{\mathbf{t}}} \text { at } \mathbf{t}=\mathbf{4} \text { is }\)
\(\begin{array}{llll}
\text { (a) } \frac{17}{4} \text { (b) } \frac{-17}{4} \text { (c) } \frac{4}{17} \text { (d) } \frac{-4}{17}
\end{array}\)

1 \(\frac{17}{4}\)
2 \(\frac{-17}{4}\)
3 \(\frac{4}{17}\)
4 \(\frac{-4}{17}\)
Application of Derivatives

85396 The equation of normal to the curve \(y=\log _{e} x\) at the point \(P(1,0)\) is

1 \(x+y=1\)
2 \(2 x+y=2\)
3 \(x-y=1\)
4 \(x-2 y=1\)
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Application of Derivatives

85393 The approximate value of \((1.002)^{\mathbf{3 0 0}}\) using differentiation is

1 1.8
2 1.6
3 1.2
4 1.4
Application of Derivatives

85394 Using differentiation, the approximate value of \(\sin 46^{\circ}\),
Given that \(1^{0}=0.0175^{\circ}\), is

1 0.07194
2 \(\frac{1.0175}{\sqrt{2}}\)
3 0.7194
4 \(\frac{0.0175}{\sqrt{2}}\)
Application of Derivatives

85395 The slope of normal to the curve \(x=\sqrt{t}\) and
\(\mathbf{y}=\mathbf{t}-\frac{1}{\sqrt{\mathbf{t}}} \text { at } \mathbf{t}=\mathbf{4} \text { is }\)
\(\begin{array}{llll}
\text { (a) } \frac{17}{4} \text { (b) } \frac{-17}{4} \text { (c) } \frac{4}{17} \text { (d) } \frac{-4}{17}
\end{array}\)

1 \(\frac{17}{4}\)
2 \(\frac{-17}{4}\)
3 \(\frac{4}{17}\)
4 \(\frac{-4}{17}\)
Application of Derivatives

85396 The equation of normal to the curve \(y=\log _{e} x\) at the point \(P(1,0)\) is

1 \(x+y=1\)
2 \(2 x+y=2\)
3 \(x-y=1\)
4 \(x-2 y=1\)